Text Formatting

danielvelizv · Post by **danielvelizv** » Fri Sep 02, 2016 1:09 am

Hello people! I have a problem, two months ago I full installed Miktex with all packages in my laptop on Windows 7, in another PC I worked a document with the same packages but when I tried to compile the same document in my laptop, it shows me the following error in the attached picture in this post, I don't know what's happening, I have the same number of packages installed on both computers. Can anyone help me on this?

$LaTeX Beginner's Guide$ $LaTeX Cookbook$ $LaTeX TikZ graphics$ $TikZによるLaTeXグラフィックス$

Post by **Johannes_B** » Fri Sep 02, 2016 8:14 am

You (or a package you are using) are using font commands that are officially obsolete since 1994. The author of KOMA-script decided to finally encourage people to stop using them.
You can use the global option enabledeprecatedfontcommands, which by definition is also deprecated.

By the way, a minimal working example and an upload of the log-file is much better than a screenshot. I cannot copy the text to run LaTeX on it myself and see the problem. Luckily, in this case the reason is easy to know.

Post by **Stefan Kottwitz** » Fri Sep 02, 2016 12:31 pm

As an addition, replace \bf by \bfseries or by \normalfont\bfseries.

Or use it like \textbf{...} similar as you already use \textcolor{...}. Such commands can be nested, such as \textbf{\textcolor{...}}.

Stefan

danielvelizv · Post by **danielvelizv** » Fri Sep 02, 2016 11:07 pm

Johannes_B wrote:You (or a package you are using) are using font commands that are officially obsolete since 1994. The author of KOMA-script decided to finally encourage people to stop using them.
You can use the global option enabledeprecatedfontcommands, which by definition is also deprecated.

By the way, a minimal working example and an upload of the log-file is much better than a screenshot. I cannot copy the text to run LaTeX on it myself and see the problem. Luckily, in this case the reason is easy to know.

Thanks buddy. I didn't know that, I was looking for this in a lot of websites, So, for both explanations I read, I have to edit all deprecated commands in my TeX file, no?

Post by **Johannes_B** » Sat Sep 03, 2016 6:31 am

That would be a good start. But some packages also use the old font commands, most notably fanyhdr. Using that package with a KOMA class is not recommended anyway, as KOMA-script comes with its own package to style header and footer. It is called scrlayer-scrpage.

danielvelizv · Post by **danielvelizv** » Sat Sep 03, 2016 7:28 pm

Johannes_B wrote:That would be a good start. But some packages also use the old font commands, most notably fanyhdr. Using that package with a KOMA class is not recommended anyway, as KOMA-script comes with its own package to style header and footer. It is called scrlayer-scrpage.

That's very good, I don't know how much I have to change my TeX. Is there any possibity to send to any of you (you or Stefan) and check and comment the modifications in my file? I want to learn about scrbook class, regards!

Post by **Stefan Kottwitz** » Sat Sep 03, 2016 7:41 pm

You can post your code here and we can make suggestions.

It's a public forum - would we start moving know-how to private messages it would not have public use anymore.

Stefan

danielvelizv · Post by **danielvelizv** » Sat Sep 03, 2016 8:12 pm

Stefan_K wrote:You can post your code here and we can make suggestions.

It's a public forum - would we start moving know-how to private messages it would not have public use anymore.

Stefan

I hope it will not be a problem on the language because I wrote in Spanish using babel, so here is my code

Code: Select all

\documentclass[10pt,letterpaper,fleqn]{scrbook}
\usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm, marginparwidth=2.85cm, marginparsep=0pt]{geometry}
\usepackage[utf8]{inputenc}
\usepackage[spanish]{babel}
\usepackage{amsmath}
\usepackage{mathtools}   %MANIPULACIÓN DE LA ALINEACIÓN LATERAL DE LAS EXPRESIONES MATEMÁTICAS
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage[most]{tcolorbox}
\usepackage{xcolor}
\usepackage{tikz}
\usepackage{array}
\usepackage{marginnote}  %COLOCACIÓN DE NOTAS DE PÁGINA EN LOS LADOS
\usepackage{setspace}    %SEPARACIÓN DE LÍNEAS EN PÁRRAFOS
\usepackage{fancyhdr}    %ENCABEZADOS DECORADOS
\usepackage{cancel}      %CANCELACIÓN DE TÉRMINOS
\usetikzlibrary{calc}
\usetikzlibrary{shapes.callouts}  %CUADROS DE IDEAS
\usetikzlibrary{decorations.text}
\usetikzlibrary{positioning}
\newcommand{\titulo}{{\fontsize{8}{0}\selectfont \sf 650 integrales indefinidas \\[-0.5ex] resueltas ¡paso a paso!}}
\newcommand{\inmediata}{{\fontsize{8}{0}\selectfont \sf Integrales inmediatas}}
\usepackage{varwidth}
\def\cabecera#1{%\x2-\x1      CABECERA EN LA PRIMERA PÁGINA DEL CAPÍTULO
 % \thispagestyle{empty}
  \begin{tikzpicture}[overlay, remember picture]
    \draw let \p1 = (current page.west), \p2 = (current page.east) in
      node[minimum width=\x2-\x1, minimum height=3cm, line width=0pt, rectangle, fill=gris!80, anchor=north west, align=left, text width=17cm] at ($(current page.north west)$)    {#1};
  \end{tikzpicture}
}
\pagestyle{fancy}

\fancyhead[LE]{\inmediata}
\fancyhead[RO]{\inmediata}
\fancyhead[RE]{\titulo}
\fancyhead[LO]{\titulo}

\fancyfoot[LE]{\bf \thepage}   %NUMERACIÓN EN LAS PÁGINAS PARES
\fancyfoot[RO]{\bf \thepage}   %NUMERACIÓN EN LAS PÁGINAS IMPARES
\fancyfoot[C]{}
\fancyfoot[RE]{\fontsize{8}{0}\selectfont \sf Ing. Daniel A. Veliz V.}
\fancyfoot[LO]{\fontsize{8}{0}\selectfont \sf Ing. Daniel A. Veliz V.} 
\renewcommand{\headrulewidth}{0.5pt}  %LÍNEA EN EL ENCABEZADO
\renewcommand{\footrulewidth}{0.5pt}  %LÍNEA EN EL PIE DE PÁGINA


\setlength{\parindent}{0pt}      %SIN SANGRÍA EN LOS PÁRRAFOS
\setlength{\arraycolsep}{4pt}    %ANCHO DE LAS COLUMNAS EN LOS ARRAY
\setlength{\tabcolsep}{4pt}      %ANCHO DE LAS COLUMNAS EN LAS TABLAS
\setlength{\mathindent}{0cm}     %SIN SANGRÍA EN LA ALINEACIÓN MATEMÁTICA
\usepackage{anyfontsize}


%---------------------------------------------
%              COLORES DEFINIDOS
%---------------------------------------------

\definecolor{naranja}{rgb}{1, 0.3, 0}
\definecolor{blanco}{rgb}{0.97, 0.97, 1}
\definecolor{gris}{rgb}{0.47, 0.53, 0.6}
\definecolor{azul}{rgb}{0.12, 0.56, 1.0}
\definecolor{verde}{rgb}{0.0, 0.65, 0.31}
\definecolor{carmin}{rgb}{1.0, 0.0, 0.22}


\begin{document}
\cabecera{\bfseries {\fontsize{20}{0}\selectfont  \hfill Capítulo I \\ \hfill Integrales inmediatas}}
\begin{minipage}[c]{1\textwidth}
\vspace*{1cm}
En este capítulo se darán a conocer los fundamentos básicos de la integración de distintas funciones por medio del empleo de las propiedades matemáticas y así convertir las funciones integrando dadas en algunas de las formas básicas presentadas antes del desarrollo de este capítulo, de esta manera a medida que revise los capitulos posteriores se dará cuenta que la idea básica de aplicar las técnicas de integración consistirá en convertir integrandos complicados en formas elementales para determinar una \emph{función primitiva} o \emph{antiderivada} de una función $f$.
\\[0.5cm]
La antiderivación (o integración indefinida) se denota mediante el signo integral     $\displaystyle \int$ por lo tanto, el siguiente esquema podrá ayudarlo a identificar los elementos implícitos en el cálculo integral y qué se obtiene al calcular una integral indefinida:
\end{minipage}
\\[0.8cm]
\hspace*{3.75cm}
\begin{tikzpicture}
\node[rectangle callout, rounded corners=3pt, draw, fill=azul!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0.7,-1)}, callout pointer width=5mm] {\begin{varwidth}{2cm} Función integrando \end{varwidth}};
\end{tikzpicture}
\hspace*{1.5cm}
\begin{tikzpicture}
\node[rectangle callout, rounded corners=3pt, draw, fill=verde!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.5,-1)}, callout pointer width=5mm] {\begin{varwidth}{2.75cm} Antiderivada de la función $f$ \end{varwidth}};
\end{tikzpicture}
\\[-0.675cm]
\begin{equation}
\hspace*{5cm} \scalebox{1.5}{$\displaystyle \int \! \textcolor{azul!100}{f(x)}\,\textcolor{naranja!90}{dx} = \textcolor{verde!100}{F(x)} \ \textcolor{carmin!100}{+ \ C}$} \nonumber
\end{equation}
\\[-0.6cm]
\hspace*{5.85cm}
\begin{tikzpicture}
\node[rectangle callout, rounded corners=3pt, draw, fill=naranja!90, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0,1)}, callout pointer width=5mm] {\begin{varwidth}{2cm}
Variable de integración \end{varwidth}};
\end{tikzpicture}
\hspace*{1.75cm}
\begin{tikzpicture}
\node[rectangle callout, rounded corners=3pt, draw, fill=carmin!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.75,1)}, callout pointer width=5mm] {\begin{varwidth}{2.5cm}
Constante de integración \end{varwidth}};
\end{tikzpicture}
\\[0.75cm]
Además, según Larson R. (2009) en su texto \emph{Cálculo Integral - Matemáticas 2} expresa que:
\vspace{1ex}
\begin{quote}
``La expresión $\displaystyle \int f(x)\,dx$ se lee como la antiderivada o primitiva de $f$ con respecto a $x$, el diferencial de $x$ sirve para identificar a $x$ como la variable de integración. El término \emph{integral indefinida} es sinónimo de antiderivada."
\end{quote}
\vspace{0.5cm}
\tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]
{\hspace*{-0.55cm} {\Huge {\bf\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{A lo largo de este texto, usted encontrará conforme vea las distintas técnicas y casos de integrandos particulares, la complejidad en el desarrollo de los mismos, como integrar cada función producirá una constante $C$, solo se asumirá en el resultado final escrito como la suma de todas las constantes de las integrales resueltas, de manera que $C = C_1 + C_2 + C_3 + \ldots + C_n$}
\vspace{0.6cm}
A continuación se presentará una lista de ejercicios con un orden aleatorio de dificultad y algunos ejemplos previamente explicados para ayudar a comprender el principio básico de la integración inmediata por medio del uso de la tabla.
\\[0.6cm]
\begin{tabular}{llllp{8cm}}
\textbf{Ej. 1.1)} $ \displaystyle \int x + 3\,dx$ & = & $\displaystyle \int x\,dx + \int 3\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[5mm]
                               & = & $\displaystyle \int x\,dx + 3 \int dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont en la integral de la derecha se extrajo el factor 3 fuera de la integral como una constante}     \\[1mm] 
               & = & $\displaystyle \frac{x^2}{2} + 3x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}
\end{tabular}
\newpage
\begin{tabular}{llllp{3cm}}
\textbf{Ej. 1.2)} $ \displaystyle \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx$ & = & $\displaystyle \int \frac{x^2}{\sqrt{x}}\,dx + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]
                               & = & $\displaystyle \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx + \int \frac{1}{x^{1/2}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont escribir los radicales en forma de potencia}    \\[1mm] 
                                & = & $\displaystyle \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]
                               & = & $\displaystyle \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar} \\[6mm]
                               & = & $\displaystyle \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}
\end{tabular}
\reversemarginpar
\marginnote{
\colorbox{yellow!30}{
\begin{minipage}{2.5cm}
\begin{spacing}{0.55}
\fontsize{7}{14}\selectfont Las funciones irracionales (raíces) cuentan como funciones de potencia.
\vspace{-8pt}
\end{spacing}
\end{minipage}}}
\\[0.9cm]
\begin{tabular}{llllp{4cm}}
\textbf{Ej. 1.3)} $ \displaystyle \int (x + 1)(3x - 2)\,dx$ & = & $\displaystyle \int 3x^2 + x - 2\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont multiplicar factores y agrupar términos semejantes} \\[1mm]
                              & = & $\displaystyle 3 \int x^2\,dx + \int x\,dx - 2 \int dx$ & $\longleftarrow$ 
&  {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]
                              & = & $\displaystyle x^3 + \frac{x^2}{2} - 2x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}
\end{tabular}
\\[0.9cm]
\begin{tabular}{llllp{4cm}}
\textbf{Ej. 1.4)} $ \displaystyle \int \sec y(\tan y - \sec y)\,dy$ & = & $\displaystyle \int \sec y \tan y - \sec^2y\,dy $ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]
                               & $=$ & $\displaystyle \int \sec y \tan y\,dy - \int \sec^2 y\,dy$ &               $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[7mm]
                               & $=$ & $\displaystyle \sec y - \tan y + C$ & $\longleftarrow$  & {\fontsize{8}{0}\selectfont integrar}
\end{tabular}
\\[0.9cm]
\begin{tabular}{llllp{6cm}}
\textbf{Ej. 1.5)} $ \displaystyle \int 2\pi y(8 - y^{3/2})\,dy$ & = & $\displaystyle 2\pi \int 8y - y^{5/2}\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont extraer el factor $2\pi$ fuera de la integral como una constante y reescribir la función}       \\[3mm]
                & = & $\displaystyle 2\pi \left[4y^2 - \frac{y^{7/2}}{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}    \\[6mm] 
                & = & $\displaystyle 2\pi \left[4y^2 - \frac{2}{7}\, y^{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}
\end{tabular}
\\
\vfill
\tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]
{\hspace*{-0.55cm} {\Huge {\bf\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{El lector observará conforme vea los ejercicios elaborados de este texto que algunos de los pasos efectuados en los ejemplos 1.1 al 1.5 en la práctica son omitidos, esto ocurrirá a medida que se familiarice con las reglas básicas de integración.}
\newpage
Calcular las siguientes integrales
\\[0.55cm]
\hspace*{-0.25cm}
%----------------------------------------------------------
%                     LISTA DE EJERCICIOS
%----------------------------------------------------------
$
{\setlength{\arraycolsep}{10pt}
\begin{array}{*3{>{\displaystyle}l}}
\textbf{1.6)} \int 2x - 3x^2\,dx & \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx & \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx \\[6mm]
\textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx  & \textbf{1.10)} \int \sqrt[3]{x^2}\,dx & \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx \\[6mm]
\textbf{1.12)} \int (2t^2 - 1)^2\,dt & \textbf{1.13)} \int y^2\sqrt{y}\,dy & \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx  \\[6mm]
\textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt & \textbf{1.16)} \int \tan^2y + 1\,dy & \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx \\[6mm]
\textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx & \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx & \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx \\[6mm]
\textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx & \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx & \textbf{1.23)} \int \frac{x^2}{x^2 + 1}\,dx \\[6mm]
\textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx & \textbf{1.25)} \int \tan^2x\,dx & \textbf{1.26)} \int \cot^2x\,dx \\[6mm]
\textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\,dx & \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx & \textbf{1.29)} \int (2^x + 3^x)^2\,dx \\[6mm]
\textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx & \textbf{1.31)} \int 3^xe^x\,dx & \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx \\[6mm]
\textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx & \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &
\textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy  \\[6mm]
\textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &   \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta & \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx \\[6mm]
\textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx & \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} & \textbf{1.41)} \int \frac{dx}{3x^2 + 5} \\[6mm]
\textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx & \textbf{1.43)} \int \left(y^2 - \frac{1}{y^2} \right)^3 dy & \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx \\[6mm]
\textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx & \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt  & \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}}
\\[6mm]
\textbf{1.48)} \int \frac{dx}{\sen x \cos x} & \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta & \textbf{1.50)} \int (\tan x + \sec x)^2\,dx  \\[6mm]
\textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx & \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx & \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi  \\[6mm]
\textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx & \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx  & \textbf{1.56)} \int 1^x\,dx
\end{array}}
$
\newpage
\hspace*{-0.35cm}
$
{\setlength{\arraycolsep}{10pt}
\begin{array}{*3{>{\displaystyle}l}}
\textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx & \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx & \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx \\[6mm]
\textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx  & \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx & \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx \\[6mm]
\textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx & \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx & \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx
\end{array}}
$
\\[1.5cm]
\textbf{\huge Solución}
\\
\rule{21cm}{1ex}
\\[1ex]

%-------------------------------------------------------------
%                           EJERCICIO 1.6
%-------------------------------------------------------------

\begin{align*}
\textbf{1.6)} \int 2x - 3x^2\,dx &= \int 2x\,dx - \int 3x^2\,dx = 2 \int x\,dx - 3 \int x^2\,dx = \cancel{2}\left(\frac{x^2}{\cancel{2}} \right) - \cancel{3} \left(\frac{x^3}{\cancel{3}} \right) + C \\[3mm]
                   &=  \fboxsep=5pt\colorbox{gris!40}{$x^2 - x^3 + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.7
%-------------------------------------------------------------

\begin{align*}
\textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx &= \int 4x^3\,dx + \int 6x^2\,dx - \int dx                									    = 4 \int x^3\,dx + 6 \int x^2\,dx - \int dx
                                       \\[3mm]
                                       &= \cancel{4} \left(\frac{x^4}{\cancel{4}} \right) + 6 \left( \frac{x^3}{3} \right) - x + C  =  \fboxsep=5pt\colorbox{gris!40}{$x^4 + 2x^3 - x + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.8
%-------------------------------------------------------------

\begin{align*}
\textbf{1.8)} \int x^{3/2} + 2x + 1\,dx &= \int x^{3/2}\,dx + 2 \int x\,dx + \int dx
										= \frac{x^{5/2}}{5/2} + \cancel{2} \left(\frac{x^2}{\cancel{2}} \right) + x + C
										\\[3mm]
				                        &=  \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{5}\,x^{5/2} + x^2 + x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.9
%-------------------------------------------------------------

\begin{align*}
\textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx &= \int \sqrt{x}\,dx + \int \frac{dx}{2\sqrt{x}} = \int x^{1/2}\,dx + \frac{1}{2} \int x^{-1/2}\,dx = \frac{x^{3/2}}{3/2} + \frac{1}{\cancel{2}} \left(\frac{x^{1/2}}{1/ \cancel{2}} \right) + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} + x^{1/2} + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.10
%-------------------------------------------------------------


\begin{align*}
\textbf{1.10)} \int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{5}\,x^{5/3} + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.11
%-------------------------------------------------------------


\begin{align*}
\textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx &= \int \frac{x^2}{x^4}\,dx + 2 \int \frac{x}{x^4}\,dx - 3 \int \frac{dx}{x^4} = \int x^{-2}\,dx + 2 \int x^{-3}\,dx - 3 \int x^{-4}\,dx		
\end{align*}
\begin{align*}
&= \frac{x^{-1}}{- 1} + \cancel{2} \left(\frac{x^{-2}}{- \cancel{2}} \right) - \cancel{3} \left(\frac{x^{-3}}{- \cancel{3}} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.12
%-------------------------------------------------------------

\begin{align*}
\textbf{1.12)} \int (2t^2 - 1)^2\,dt &= \int 4t^4 - 4t^2 + 1\,dt = 4 \int t^4\,dt - 4 \int t^2\,dt + \int dt = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{4}{5} \, t^5 - \frac{4}{3} \, t^3 + t + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.13
%-------------------------------------------------------------

\begin{align*}
\textbf{1.13)} \int y^2\sqrt{y}\,dy = \int y^{5/2}\,dy = \frac{y^{7/2}}{7/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{7}\,y^{7/2} + C$}  
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.14
%-------------------------------------------------------------

\begin{align*}
\textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx = 2 \int \sen x\,dx + 3 \int \cos x\,dx = \fboxsep=5pt\colorbox{gris!40}{$- 2 \cos x + 3 \sen x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.15
%-------------------------------------------------------------

\begin{align*}
\textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt &= \int \frac{1 - t(t^2 + 1)}{t^2 + 1} = \int \frac{dt}{t^2 + 1} - \int \frac{t (\cancel{t^2 + 1})}{\cancel{t^2 + 1}}\,dt = \fboxsep=5pt\colorbox{gris!40}{$\arctan t - \displaystyle \frac{t^2}{2} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.16
%-------------------------------------------------------------

\begin{align*}
\textbf{1.16)} \int \tan^2y + 1\,dy = \int \sec^2y - 1 + 1\,dy = \fboxsep=5pt\colorbox{gris!40}{$\tan y + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.17
%-------------------------------------------------------------

\begin{align*}
\textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx &= \int x^2\,dx + \int \frac{dx}{9x^2}  					       = \int x^2\,dx + \frac{1}{9} \int x^{-2}\,dx               						   = \frac{x^3}{3} + \frac{1}{9}\left(\frac{x^{-1}}{-1} 	 							   \right) + C \\[3mm]
                           &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 									   \frac{x^3}{3} - \frac{1}{9x} + C $}  
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.18
%-------------------------------------------------------------


\begin{align*}
\textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx &= \int \left(1 - \frac{1}{x^2} \right) \left(x^{3/2} \right)^{1/2}\,dx = \int \left(1 - \frac{1}{x^2} \right)x^{3/4}\,dx \\[3mm]
&= \int x^{3/4}\,dx - \int x^{-5/4}\,dx = \frac{x^{7/4}}{7/4} - \left(- \frac{x^{-1/4}}{1/4} \right) + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{4}{7}\,x^{7/4} + \frac{4}{x^{1/4}} + C $}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.19
%-------------------------------------------------------------

\begin{align*}
\textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx &= \int \frac{2x - 2\sqrt{2x}\sqrt[3]{3x} + {(\sqrt[3]{3x})}^2}{x}\,dx \\[3mm]
&= 2 \int dx - 2 \int \frac{\sqrt{2}\,x^{1/2}\,\sqrt[3]{3}\,{x}^{1/3}}{x}\,dx + \int \frac{\sqrt[3]{9}\,x^{2/3}}{x}\,dx \\[3mm]
&= 2 \int dx - 2 \sqrt{2} \sqrt[3]{3} \int x^{-1/6}\,dx + \sqrt[3]{9} \int x^{-1/3}\,dx \\[3mm]
&= 2x - 2 \sqrt{2} \sqrt[3]{3} \left(\frac{x^{5/6}}{5/6} \right) + \sqrt[3]{9} \left( \frac{x^{2/3}}{2/3} \right) + C
\end{align*}
\begin{align*}
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x - \frac{12}{5} \sqrt{2} \sqrt[3]{3}\,x^{5/6} + \frac{3}{2} \sqrt[3]{9}\,x^{2/3} + C $}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.20
%-------------------------------------------------------------

\begin{align*}
\textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx &= \int \sqrt[3]{\sqrt{\frac{2x^2}{x}}}\,dx = \int \sqrt[3]{\sqrt{2x}}\,dx  = \int {\left[(2x)^{1/2} \right]}^{1/3}\,dx = \int (2x)^{1/6}\,dx  \\[3mm]
&= \sqrt[6]{2} \left(\frac{x^{7/6}}{7/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt[6]{2}}{7}\,x^{7/6} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.21
%-------------------------------------------------------------

\begin{align*}
\textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx &= \int \frac{2^x\,2}{10^x}\,dx - \int \frac{5^x}{5\,10^x}\,dx = 2 \int \left(\frac{2}{10} \right)^x\,dx - \frac{1}{5} \int \left(\frac{5}{10} \right)^x\,dx \\[3mm]
&= 2 \left[ \frac{(1/5)^x}{\ln(1/5)} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln(1/2)} \right] + C = 2 \left[ \frac{(1/5)^x}{\ln 1 - \ln 5} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln 1- \ln 2} \right] + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{5 \ln 2} \left(\frac{1}{2} \right)^x - \frac{2}{\ln 5} \left(\frac{1}{5} \right)^x + C $}
\end{align*}
%\reversemarginpar
\marginnote{
\colorbox{yellow!30}{
\begin{minipage}{2.5cm}
\begin{spacing}{0.55}
\fontsize{7}{14}\selectfont Es menester destacar que en el ejercicio 1.23 no era necesario aplicar el algoritmo de la división, simplemente con sumar y restar el factor 1 en el numerador y separar las fracciones se obtendría el mismo resultado, este artificio de sumar y restar, multiplicar y dividir elementos será de gran utilidad para la resolución de un gran número de ejercicios presentados en este texto.
\vspace{-8pt}
\end{spacing}
\end{minipage}}
}

%-------------------------------------------------------------
%                          EJERCICIO 1.22
%-------------------------------------------------------------

\begin{align*}
\textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx &= \int \frac{\sqrt{x^4 + \displaystyle \frac{1}{x^4} + 2}}{x^3}\,dx = \int \frac{\displaystyle \sqrt{\frac{x^8 + 2x^4 + 1}{x^4}}}{x^3}\,dx \\[3mm]
&= \int \frac{\sqrt{x^8 + 2x^4 + 1}}{x^5}\,dx = \int \frac{\sqrt{(x^4 + 1)^2}}{x^5}\,dx = \int \frac{x^4 + 1}{x^5}\,dx  \\[3mm]
&= \int \frac{dx}{x} + \int x^{-5}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| x \big| - \frac{1}{4x^4} + C$}
\end{align*}
\vspace{1cm}


%-------------------------------------------------------------
%                          EJERCICIO 1.23
%-------------------------------------------------------------


\textbf{1.23)} $ \displaystyle \int \frac{x^2}{x^2 + 1}\,dx$ \quad al aplicar la división de polinomios
\quad
$
\begin{array}{cccc|ccc}
   & \cancel{x^2} & + & 0 & x^2 & + & 1 \\
\cline{5-7}
 - & \cancel{x^2} & - & 1 & 1   &   &   \\
\cline{2-4}  
   &              & - & 1 &     &   &
\end{array}
$
\\[0.25cm]
\begin{align*}
\mbox{La integral se convierte en} \int \frac{x^2}{x^2 + 1}\,dx &= \int \frac{(x^2 + 1)1 - 1}{x^2 + 1}\,dx = \int \frac{\cancel{x^2 + 1}}{\cancel{x^2 + 1}}\,dx - \int \frac{dx}{x^2 + 1}       \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle x - \arctan x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.24
%-------------------------------------------------------------

\begin{align*}
\textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx &= \int \frac{\cancel{(e^x + 1)}(e^{2x} - e^x + 1)}{\cancel{e^x + 1}}\,dx = \int e^{2x} - e^x + 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{e^{2x}}{2} - e^x + x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.25
%-------------------------------------------------------------

\begin{align*}
\textbf{1.25)} \int \tan^2x\,dx = \int \sec^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\tan x - x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.26
%-------------------------------------------------------------


\begin{align*}
\textbf{1.26)} \int \cot^2x\,dx = \int \csc^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$- \cot x - x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.27
%-------------------------------------------------------------


\begin{align*}
\textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx &= \int \frac{\sqrt{1 + x^2}}{\sqrt{1 - x^4}}\,dx + \int \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx \\[3mm]
&= \int \sqrt{\frac{\cancel{1 + x^2}}{(1 - x^2)\cancel{(1 + x^2)}}}\,dx + \int \sqrt{\frac{\cancel{1 - x^2}}{\cancel{(1 - x^2)}(1 + x^2)}}\,dx \\[3mm]
&= \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{dx}{\sqrt{x^2 + 1}} \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\arcsen x + \ln(x + \sqrt{x^2 + 1}) + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.28
%-------------------------------------------------------------

\begin{align*}
\textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx &= \int \frac{1 - 3x^2 + 3x - x^3}{x\,x^{1/3}}\,dx
\\[3mm]
&= \int x^{-4/3}\,dx - 3 \int x^{2/3}\,dx + 3 \int x^{-1/3}\,dx - \int x^{5/3}\,dx \\[3mm]
&= \frac{x^{-1/3}}{-1/3} - 3 \, \frac{x^{5/3}}{5/3} + 3 \, \frac{x^{2/3}}{2/3} - \frac{x^{8/3}}{8/3} + C
\\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{3}{x^{1/3}} - \frac{9}{5}\,x^{5/3} + \frac{9}{2}\,x^{2/3} - \frac{3}{8}\,x^{8/3} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.29
%-------------------------------------------------------------

\begin{align*}
\textbf{1.29)} \int (2^x + 3^x)^2\,dx &= \int {(2^x)}^2 + 2(2^x)(3^x) + {(3^x)}^2\,dx = \int 4^x\,dx + 2 \int 6^x\,dx + \int 9^x\,dx \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln 4}\,4^x + \frac{2}{\ln 6}\,6^x + \frac{1}{\ln 9}\,9^x + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.30
%-------------------------------------------------------------

\begin{align*}
\textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx &= \int n^{\frac{1 - n}{n}}\,x^{\frac{1 - n}{n}}\,dx
= n^{\frac{1 - n}{n}} \int x^{\frac{1 - n}{n}}\,dx = n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - n}{n} + 1}}{\frac{1 - n}{n} + 1} \right) + C \\[2mm]
&= n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - \cancel{n} + \cancel{n}}{n}}}{\frac{1 - \cancel{n} + \cancel{n}}{n}} \right) + C = n^{\frac{1 - n}{n}}\,n\,x^{1/n} + C = n^{1/n}\,x^{1/n} + C \\[2mm]
&= (nx)^{1/n} + C = \fboxsep=5pt\colorbox{gris!40}{$\sqrt[n]{nx} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.31
%-------------------------------------------------------------


\begin{align*}
\textbf{1.31)} \int 3^xe^x\,dx &= \int (3e)^x\,dx = \frac{(3e)^x}{\ln(3e)} + C = \frac{(3e)^x}{\ln 3 + \ln e} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3^xe^x}{\ln 3 + 1} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.32
%-------------------------------------------------------------


\begin{align*}
\textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx &= \int {(a^{2/3})}^3 - 3{(a^{2/3})}^2x^{2/3} + 3a^{2/3}{(x^{2/3})}^2 - {(x^{2/3})}^3\,dx \\[3mm]
&= \int a^2 - 3a^{4/3}x^{2/3} + 3a^{2/3}x^{4/3} - x^2\,dx \\[3mm]
&= a^2 \int dx - 3a^{4/3} \int x^{2/3}\,dx + 3a^{2/3} \int x^{4/3}\,dx - \int x^2\,dx \\[3mm]
&= a^2x - 3a^{4/3}\left(\frac{x^{5/3}}{5/3} \right) + 3a^{2/3} \left(\frac{x^{7/3}}{7/3} \right) - \frac{x^3}{3} + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle a^2x - \frac{9a^{4/3}}{5}\,x^{5/3} + \frac{9a^{2/3}}{7}\,x^{7/3} - \frac{x^3}{3} + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.33
%-------------------------------------------------------------

\begin{align*}
\textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx &= \int \frac{x^{2m} - 2x^mx^n + x^{2n}}{\sqrt{x}}\,dx  \\[3mm]
&= \int x^{2m - 1/2}\,dx - 2 \int x^{m + n - 1/2}\,dx + \int x^{2n - 1/2}\,dx \\[3mm]
&= \frac{x^{2m - 1/2 + 1}}{2m - 1/2 + 1} - 2 \left(\frac{x^{m + n - 1/2 + 1}}{m + n - 1/2 + 1} \right) + \frac{x^{2n - 1/2 + 1}}{2n - 1/2 + 1} + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2 \left( \frac{x^{\frac{4m + 1}{2}}}{4m + 1} \right) - 4 \left( \frac{x^{\frac{2m + 2n + 1}{2}}}{2m + 2n + 1} \right) + 2 \left( \frac{x^{\frac{4n + 1}{2}}}{4n + 1} \right) + C$}
\end{align*}

%-------------------------------------------------------------
%                          EJERCICIO 1.34
%-------------------------------------------------------------

\begin{align*}
\textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &= \int \frac{a^{2x} - 2a^xb^x + b^{2x}}{a^xb^x}\,dx = \int \frac{a^{2x}}{a^xb^x}\,dx -2 \int \frac{\cancel{a^xb^x}}{\cancel{a^xb^x}}\,dx + \int \frac{b^{2x}}{a^xb^x}\,dx \\[2mm]
&= \int \left(\frac{a}{b} \right)^xdx - 2 \int dx + \int \left(\frac{b}{a} \right)^xdx \\[2mm]
&= \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln (a/b)} - 2x + \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln (b/a)} + C = \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln a - \ln b} - 2x - \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln a - \ln b} + C \\[2mm]
&= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{1}{\ln a - \ln b} \left[ \left(\frac{a}{b} \right)^x - \left(\frac{b}{a} \right)^x \right] - 2x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.35
%-------------------------------------------------------------


\begin{align*}
\textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy &= 4 \int y^3\,dy + 2 \int \frac{dy}{y^3} = \cancel{4}\left(\frac{y^4}{\cancel{4}} \right) - \cancel{2}\left(\frac{1}{- \cancel{2}y^2} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle y^4 + \frac{1}{y^2} + C $}
\end{align*}
\vspace{-0.65cm}


%-------------------------------------------------------------
%                          EJERCICIO 1.36
%-------------------------------------------------------------


\begin{align*}
\textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &= \int \left(\frac{1}{\sqrt{2}\sen x} \right)^2 - 2\left(\frac{1}{\sqrt{2} \sen x} \right) + 1\,dx \\[3mm]
&= \int \frac{dx}{2 \sen^2x} - 2 \int \frac{dx}{\sqrt{2} \sen x} + \int dx = \frac{1}{2} \int \csc^2x\,dx - \sqrt{2} \int \csc x\,dx  + \int dx \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2} \cot x - \sqrt{2} \ln \big|\csc x - \cot x \big| + x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.37
%-------------------------------------------------------------


\begin{align*}
\textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta(1 + \tan^2\theta)}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta\,\cancel{\sec^2\theta}}{\cancel{\sec^2\theta}}\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \cos \theta + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.38
%-------------------------------------------------------------


\begin{align*}
\textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx &= \int \frac{\sen x}{\cos^2x}\,dx = \int \frac{1}{\cos x}\,\frac{\sen x}{\cos x}\,dx = \int \sec \theta\,\tan \theta\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \sec \theta + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.39
%-------------------------------------------------------------


\begin{align*}
\textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx &= \int \frac{2\,\cancel{\sen x}\,\cos x}{\cancel{\sen x}}\,dx = \fboxsep=5pt\colorbox{gris!40}{$2\,\sen x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.40
%-------------------------------------------------------------


\begin{align*}
\textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} &= \int \frac{dx}{(a - b) \left[\displaystyle \frac{a + b}{a - b} - x^2 \right]} = \frac{1}{a - b} \int \frac{dx}{\left(\sqrt{\frac{a + b}{a - b}} \right)^2 - x^2} \\[3mm]
&= \frac{1}{2(a - b)\displaystyle \sqrt{\frac{a + b}{a - b}}} \ln \left|\displaystyle \frac{\displaystyle \sqrt{\frac{a + b}{a - b}} + x}{\displaystyle \sqrt{\frac{a + b}{a - b}} - x} \right| + C \\[3mm]
&= \frac{1}{2\sqrt{(a + b)(a - b)}} \ln \left| \frac{\displaystyle  \frac{\sqrt{a + b} \ + \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}}{\displaystyle  \frac{\sqrt{a + b} \ - \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}} \right| + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{2 \sqrt{a^2 - b^2}} \ln \left| \frac{\sqrt{a + b} + \sqrt{a - b}x}{\sqrt{a + b} - \sqrt{a - b}x} \right| + C$}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.41
%-------------------------------------------------------------



\begin{align*}
\textbf{1.41)} \int \frac{dx}{3x^2 + 5} &= \int \frac{dx}{3\left(x^2 + \frac{5}{3} \right)}
= \frac{1}{3} \int \frac{dx}{x^2 + \left(\sqrt{\frac{5}{3}} \right)^2} = \frac{1}{3} \left[\frac{1}{\sqrt{\frac{5}{3}}} \arctan \left(\frac{x}{\sqrt{\frac{5}{3}}} \right) \right] + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{3}}{3\sqrt{5}} \arctan \left(\frac{\sqrt{3} x}{\sqrt{5}} \right) + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.42
%-------------------------------------------------------------


\begin{align*}
\textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx &= \int {(\sqrt{x})}^2 - 2 \left(\cancel{\sqrt{x}} \frac{1}{\cancel{\sqrt{x}}} \right) + \left(\frac{1}{\sqrt{x}} \right)^2 dx = \int x^2\,dx - 2 \int dx + \int \frac{dx}{x} \\[3mm]
& = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - 2x + \ln \big| x \big| + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.43
%-------------------------------------------------------------


\begin{align*}
\textbf{1.43)}\int \left(y^2 - \frac{1}{y^2} \right)^3 dy &= \int {(y^2)}^3 - 3{(y^2)}^2\left(\frac{1}{y^2} \right) + 3y^2 \left(\frac{1}{y^2} \right)^2 - \left(\frac{1}{y^2} \right)^{\!\! 3} dy \\[3mm]
&= \int y^6\,dy - 3 \int y^2\,dy + 3 \int y^{-2}\,dy - \int y^{-6}\,dy \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{y^7}{7} - y^3 - \frac{3}{y} + \frac{1}{5y^5} + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.44
%-------------------------------------------------------------


\begin{align*}
\textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx &= \int \left( e^{x/a} \right)^2 - 2 e^{x/a} e^{- x/a} + \left( e^{- x/a} \right)^2 dx \\[3mm]
&= \int e^{2x/a} dx - 2 \int e^0\,dx + \int e^{-2x/a} dx \\[3mm]
&= \frac{e^{2x/a}}{\frac{2}{a}} - 2x - \frac{e^{- 2x/a}}{\frac{2}{a}} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{a}{2} \left[e^{2x/a} - e^{- 2x/a} \right] - 2x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.45
%-------------------------------------------------------------


\begin{align*}
\textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx &= \frac{\sqrt{5}}{5} \int x^{1/2}\,dx + \sqrt{5} \int x^{-1/2}\,dx = \frac{\sqrt{5}}{5} \left( \frac{2}{3} x^{3/2} \right) + \sqrt{5}(2 \sqrt{x}) + C \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2\sqrt{5}}{15}\,x^{3/2} + 2 \sqrt{5x} + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.46
%-------------------------------------------------------------


\begin{align*}
\textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt = 4 \int {(e^t)}^{-1/2} = - 4 \, \frac{e^{-t/2}}{1/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{8}{\sqrt{e^t}} + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.47
%-------------------------------------------------------------


%\vspace*{-1cm}
\begin{align*}
\textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} &= \int \frac{dx}{\sqrt{5\left(\frac{7}{5} - x^2 \right)}} = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(\sqrt{\frac{7}{5}} \right)^2 - x^2}} = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{\sqrt{5}}{5} \arcsen \left(\frac{\sqrt{5}x}{\sqrt{7}} \right) + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.48
%-------------------------------------------------------------


\begin{align*}
\textbf{1.48)} \int \frac{dx}{\sen x \cos x} &= \int \frac{\sen^2x + \cos^2x}{\sen x \cos x}\,dx = \int \frac{\sen^2x}{\sen x \cos x}\,dx + \int \frac{\cos^2x}{\sen x \cos x}\,dx = \int \tan x\,dx + \\[3mm]
& \int \cot x\,dx = - \ln \big| \cos x \big| + \ln \big| \sen x \big| + C = \ln \left| \frac{\sen x}{\cos x} \right| + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \ln \big| \tan x \big | + C$} 
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.49
%-------------------------------------------------------------


\begin{align*}
\textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta = \int \frac{d \theta}{\cos^2\theta} + \int \frac{\cancel{\cos^2 \theta}}{\cancel{\cos^2 \theta}}\,d \theta = \int \sec^2 \theta\,d \theta + \int d \theta
= \fboxsep=5pt\colorbox{gris!40}{$ \tan \theta + \theta + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.50
%-------------------------------------------------------------


\begin{align*}
\textbf{1.50)} \int (\tan x + \sec x)^2\,dx &= \int \tan^2x + 2 \tan x\,\sec x + \sec^2x\,dx 
\\[3mm]
&= \int \sec^2x - 1 + 2 \tan x\,\sec x + \sec^2x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2\tan x + 2\sec x - x + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.51
%-------------------------------------------------------------


\begin{align*}
\textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx = \int \left(x^{3/2} \right)^{1/2}dx = \int x^{3/4}dx = \frac{x^{7/4}}{7/4} + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{4}{7}\,x^{7/4} + C$}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.52
%-------------------------------------------------------------


\begin{align*}
\textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx &= \int \frac{(3x^3 - 2) \cancel{(3x^3 + 2)}}{\cancel{(3x^3 + 2)}}\,dx = \int 3x^3\,dx - \int 2\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{3}{4}\,x^4 - 2x + C$}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.53
%-------------------------------------------------------------


\begin{align*}
\textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi &= \int \frac{\displaystyle \frac{1}{\sen \phi}}{\displaystyle \frac{1}{\sen \phi} - \sen \phi}\,d\phi = \int \frac{\displaystyle \frac{1}{\cancel{\sen \phi}}}{\displaystyle \frac{1 - \sen^2\phi}{\cancel{\sen \phi}}}\,d \phi = \int \frac{1}{\cos^2\phi}\,d\phi = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \tan \phi + C$}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.54
%-------------------------------------------------------------


\marginnote{
\colorbox{yellow!30}{
\begin{minipage}{2.5cm}
\begin{spacing}{0.55}
\fontsize{7}{14}\selectfont Recuerde que la expresión $\int \frac{\sen x}{\cos^2x}\,dx$ ya había aparecido antes en el ejercicio 1.38 por lo que se omitieron algunos detalles de solución.
\vspace{-8pt}
\end{spacing}
\end{minipage}}}
\vspace*{-0.25cm}
\begin{align*}
\textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx &= \int \frac{\sen x}{\cos x}\,dx + \int \frac{\tan x}{\cos x}\,dx = \int \tan x\,dx + \int \frac{\sen x}{\cos^2x}\,dx \\[3mm]
&= \int \tan x\,dx + \int \sec x \tan x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle - \ln \big| \cos x \big| + \sec x + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.55
%-------------------------------------------------------------

\marginnote{
\colorbox{yellow!30}{ 
\begin{minipage}{2.5cm}
\begin{spacing}{0.55}
\fontsize{7}{14}\selectfont La integral del ejercicio 1.56 es de la forma $a^x$ pero $a = 1$ por lo que no se puede aplicar la forma de la tabla, pero si las propiedades de los logaritmos.
\vspace{-8pt}
\end{spacing}
\end{minipage}}}


\begin{align*}
\textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx &= \frac{\sqrt{x(x - 1)} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx \\[3mm]
&= \int \frac{\sqrt{x}\sqrt{x - 1} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx = \int \frac{\cancel{\sqrt{x - 1}}(\sqrt{x} - e^x)}{\cancel{\sqrt{x - 1}}}\,dx \\[3mm]
&= \int \sqrt{x}\,dx - \int e^x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{2}{3}\,x^{3/2} - e^x + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.56
%-------------------------------------------------------------


\begin{align*}
\textbf{1.56)} \int 1^x\,dx = \int e^{\ln 1^x}dx = \int e^{x\,\ln 1}dx = \int e^0dx = \int dx = \fboxsep=5pt\colorbox{gris!40}{$ x + C$}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.57
%-------------------------------------------------------------


\begin{align*}
\textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx &= \int x^{5/3}x^{-4/3}(x^3 - 1)\,dx = \int x^{1/3}(x^3 - 1)\,dx  \\[3mm]
&= \int x^{10/3}dx - \int x^{1/3}dx = \frac{x^{13/3}}{13/3} + \frac{x^{4/3}}{4/3} + C 
\end{align*}
\begin{align*}
= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{13}\,x^{13/3} + \frac{3}{4}\,x^{4/3} + C $}
\end{align*}



%-------------------------------------------------------------
%                          EJERCICIO 1.58
%-------------------------------------------------------------


\begin{align*}
\textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx &= \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{(\sqrt{2x} - \sqrt{x})(\sqrt{2x} + \sqrt{x})}\,dx = \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{2x - x}\,dx
\\[3mm]
&= \int \frac{\sqrt{2}x^{3/2} + x^{3/2} - \sqrt{2}x^{1/2} - x^{1/2}}{x}\,dx \\[3mm]
&= \sqrt{2} \int x^{1/2}\,dx + \int x^{1/2}\,dx - \sqrt{2} \int x^{-1/2}\,dx - \int x^{- 1/2}\,dx \\[3mm]
&= ( \sqrt{2} + 1) \int x^{1/2}\,dx - (\sqrt{2} + 1) \int x^{-1/2}\,dx \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle (\sqrt{2} + 1) \left[\frac{2}{3}\,x^{3/2} - 2x^{1/2} \right] + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.59
%-------------------------------------------------------------


\begin{align*}
\textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx &= \frac{\sqrt{5}}{\sqrt[3]{3}} \int \frac{x^{3/2}}{x^{1/3}}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \int x^{7/6}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \left( \frac{x^{13/6}}{13/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt{5}}{13 \sqrt[3]{3}}\,x^{13/6} + C$}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.60
%-------------------------------------------------------------


\begin{align*}
\textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx &= \int \frac{\tan x}{3 \sen x}\,dx - \int \frac{\sen^2x}{3 \sen x}\,dx + \int \frac{4 \cos x}{3 \sen x}\,dx \\[3mm]
&= \frac{1}{3} \int \frac{\cancel{\sen x}}{\cos x \, \cancel{\sen x}}\,dx - \frac{1}{3} \int \sen x\,dx + \frac{4}{3} \int \cot x\,dx \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{1}{3}\,\ln \big| \sec x + \tan x \big| + \frac{1}{3} \cos x + \frac{4}{3} \ln \big| \sen x \big| + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.61
%-------------------------------------------------------------


\begin{align*}
\textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx &= \int \left(\frac{1}{x} \right)^3 - 3 \left(\frac{1}{x} \right)^2 x + 3 \left(\frac{1}{x} \right)\,x^2 - x^3\,dx \\[3mm]
&= \int x^{-3}\,dx - 3 \int \frac{dx}{x} + 3 \int x\,dx - \int x^3\,dx \\[3mm]
&= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle - \frac{1}{2\,x^2} - 3\,\ln \big| x \big| + \frac{3}{2}\,x^2 - \frac{x^4}{4} + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.62
%-------------------------------------------------------------


\begin{align*}
\textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx = \int \frac{\cancel{e^x}\,e^2}{\cancel{e^x}\,e}\,dx = \int e\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle ex + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.63
%-------------------------------------------------------------


\begin{align*}
\textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx = \int 8x^3 - 6x^2 - x^{-3}\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2x^4 - 2x^3 + \frac{1}{2x^2} + C $}
\end{align*}


%-------------------------------------------------------------
%                          EJERCICIO 1.64
%-------------------------------------------------------------


\begin{align*}
\textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx = \int \frac{4 \, \cancel{\ln x}}{\cancel{\ln x}}\,dx = 4 \int dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 4x + C $}
\end{align*}
\marginnote{
\colorbox{yellow!30}{
\begin{minipage}{2.5cm}
\begin{spacing}{0.55}
\fontsize{7}{14}\selectfont Observe con cuidado que en el ejercicio 1.64, la función logaritmo del numerador \emph{no está elevada a la cuarta potencia}, solo su argumento, por eso fue posible la aplicación de la propiedad y la posterior la simplificación de los términos.
\vspace{-8pt}
\end{spacing}
\end{minipage}}}


%-------------------------------------------------------------
%                          EJERCICIO 1.65
%-------------------------------------------------------------


\vspace*{-0.8cm}
\begin{align*}
\textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx &= \int \frac{1 - \sen x}{(1 + \sen x)(1 - \sen x)}\,dx = \int \frac{1 - \sen x}{1 - \sen^2x}\,dx 
= \int \frac{1 - \sen x}{\cos^2x}\,dx
\\[3mm]
&= \int \sec^2x\,dx - \int \frac{\sen x}{\cos^2x}\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \tan x - \sec x + C $}
\end{align*}
\end{document}

Post by **Johannes_B** » Sun Sep 04, 2016 11:45 am

That's the good thing, the relevant LaTeX code is the same for all of us. There is no language barrier.

Another thing though, you really should have read soma basics about LaTeX. You are doing everything by hand, leaving behind spaghetti code of the worst kind.

I'll see what i can do to slim things up a bit.

Post by **Johannes_B** » Sun Sep 04, 2016 12:07 pm

The following at least took care of the headers and footers. The rest, it is just too much to invest time in it.

For your next projects, use proper empty lines to separate paragraphs. You are making LaTeX a very hard time by avoiding them.
Don't use \displaystyle in inline mode. It make the text ugly.
Don't number stuff by hand. LaTeX is supposed to do that for you.
There are excellent pacages like exsheets that help you setting up this kind of documents. Also very helpful for what you want to do: tasks.

Code: Select all

    \documentclass[10pt,letterpaper,fleqn,
    headsepline,footsepline,
    plainheadsepline,plainfootsepline,
    ]{scrbook}
    \usepackage[left=3cm,right=2.5cm,top=2.5cm,bottom=2.5cm, marginparwidth=2.85cm, marginparsep=0pt,head=22.22223pt]{geometry}
    \usepackage[utf8]{inputenc}
    \usepackage[spanish]{babel}
    \usepackage{amsmath}
    \usepackage{mathtools}   %MANIPULACIÓN DE LA ALINEACIÓN LATERAL DE LAS EXPRESIONES MATEMÁTICAS
    \usepackage{amsfonts}
    \usepackage{amssymb}
    \usepackage{graphicx}
    \usepackage[most]{tcolorbox}
    \usepackage{xcolor}
    \usepackage{tikz}
    \usepackage{array}
    \usepackage{marginnote}  %COLOCACIÓN DE NOTAS DE PÁGINA EN LOS LADOS
    \usepackage{setspace}    %SEPARACIÓN DE LÍNEAS EN PÁRRAFOS
%    \usepackage{fancyhdr}    %ENCABEZADOS DECORADOS
    \usepackage{cancel}      %CANCELACIÓN DE TÉRMINOS
    \usetikzlibrary{calc}
    \usetikzlibrary{shapes.callouts}  %CUADROS DE IDEAS
    \usetikzlibrary{decorations.text}
    \usetikzlibrary{positioning}
    \usepackage{varwidth}
    \def\cabecera#1{%\x2-\x1      CABECERA EN LA PRIMERA PÁGINA DEL CAPÍTULO
     % \thispagestyle{empty}
      \begin{tikzpicture}[overlay, remember picture]
        \draw let \p1 = (current page.west), \p2 = (current page.east) in
          node[minimum width=\x2-\x1, minimum height=3cm, line width=0pt, rectangle, fill=gris!80, anchor=north west, align=left, text width=17cm] at ($(current page.north west)$)    {#1};
      \end{tikzpicture}
    }
%    \pagestyle{fancy}
%
%    \fancyhead[LE]{\inmediata}
%    \fancyhead[RO]{\inmediata}
%    \fancyhead[RE]{\titulo}
%    \fancyhead[LO]{\titulo}
%
%    \fancyfoot[LE]{\bf \thepage}   %NUMERACIÓN EN LAS PÁGINAS PARES
%    \fancyfoot[RO]{\bf \thepage}   %NUMERACIÓN EN LAS PÁGINAS IMPARES
%    \fancyfoot[C]{}
%    \fancyfoot[RE]{\fontsize{8}{0}\selectfont \sf Ing. Daniel A. Veliz V.}
%    \fancyfoot[LO]{\fontsize{8}{0}\selectfont \sf Ing. Daniel A. Veliz V.}
%    \renewcommand{\headrulewidth}{0.5pt}  %LÍNEA EN EL ENCABEZADO
%    \renewcommand{\footrulewidth}{0.5pt}  %LÍNEA EN EL PIE DE PÁGINA

    \newcommand{\titulo}{{650 integrales indefinidas \\ resueltas ¡paso a paso!}}
    \newcommand{\inmediata}{{Integrales inmediatas}}
    \usepackage{scrlayer-scrpage}
    \addtokomafont{pagenumber}{\bfseries}
    \addtokomafont{pageheadfoot}{\fontsize{8}{9}\sffamily\upshape}
    \clearpairofpagestyles
    \ohead*{\inmediata}
    \ihead*{\titulo}
    \ofoot*{\pagemark}
    \ifoot*{Ing. Daniel A. Veliz V.}

    \setlength{\parindent}{0pt}      %SIN SANGRÍA EN LOS PÁRRAFOS
    \setlength{\arraycolsep}{4pt}    %ANCHO DE LAS COLUMNAS EN LOS ARRAY
    \setlength{\tabcolsep}{4pt}      %ANCHO DE LAS COLUMNAS EN LAS TABLAS
    \setlength{\mathindent}{0cm}     %SIN SANGRÍA EN LA ALINEACIÓN MATEMÁTICA
    \usepackage{anyfontsize}


    %---------------------------------------------
    %              COLORES DEFINIDOS
    %---------------------------------------------

    \definecolor{naranja}{rgb}{1, 0.3, 0}
    \definecolor{blanco}{rgb}{0.97, 0.97, 1}
    \definecolor{gris}{rgb}{0.47, 0.53, 0.6}
    \definecolor{azul}{rgb}{0.12, 0.56, 1.0}
    \definecolor{verde}{rgb}{0.0, 0.65, 0.31}
    \definecolor{carmin}{rgb}{1.0, 0.0, 0.22}


    \begin{document}
    \cabecera{\bfseries {\fontsize{20}{0}\selectfont  \hfill Capítulo I \\ \hfill Integrales inmediatas}}
    \begin{minipage}[c]{1\textwidth}
    \vspace*{1cm}
    En este capítulo se darán a conocer los fundamentos básicos de la integración de distintas funciones por medio del empleo de las propiedades matemáticas y así convertir las funciones integrando dadas en algunas de las formas básicas presentadas antes del desarrollo de este capítulo, de esta manera a medida que revise los capitulos posteriores se dará cuenta que la idea básica de aplicar las técnicas de integración consistirá en convertir integrandos complicados en formas elementales para determinar una \emph{función primitiva} o \emph{antiderivada} de una función $f$.
    \\[0.5cm]
    La antiderivación (o integración indefinida) se denota mediante el signo integral     $\displaystyle \int$ por lo tanto, el siguiente esquema podrá ayudarlo a identificar los elementos implícitos en el cálculo integral y qué se obtiene al calcular una integral indefinida:
    \end{minipage}
    \\[0.8cm]
    \hspace*{3.75cm}
    \begin{tikzpicture}
    \node[rectangle callout, rounded corners=3pt, draw, fill=azul!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0.7,-1)}, callout pointer width=5mm] {\begin{varwidth}{2cm} Función integrando \end{varwidth}};
    \end{tikzpicture}
    \hspace*{1.5cm}
    \begin{tikzpicture}
    \node[rectangle callout, rounded corners=3pt, draw, fill=verde!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.5,-1)}, callout pointer width=5mm] {\begin{varwidth}{2.75cm} Antiderivada de la función $f$ \end{varwidth}};
    \end{tikzpicture}
    \\[-0.675cm]
    \begin{equation}
    \hspace*{5cm} \scalebox{1.5}{$\displaystyle \int \! \textcolor{azul!100}{f(x)}\,\textcolor{naranja!90}{dx} = \textcolor{verde!100}{F(x)} \ \textcolor{carmin!100}{+ \ C}$} \nonumber
    \end{equation}
    \\[-0.6cm]
    \hspace*{5.85cm}
    \begin{tikzpicture}
    \node[rectangle callout, rounded corners=3pt, draw, fill=naranja!90, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(0,1)}, callout pointer width=5mm] {\begin{varwidth}{2cm}
    Variable de integración \end{varwidth}};
    \end{tikzpicture}
    \hspace*{1.75cm}
    \begin{tikzpicture}
    \node[rectangle callout, rounded corners=3pt, draw, fill=carmin!100, inner sep=0.1cm, column sep=0.3cm, minimum width=0.5cm, minimum height=0.5cm, callout relative pointer={(-0.75,1)}, callout pointer width=5mm] {\begin{varwidth}{2.5cm}
    Constante de integración \end{varwidth}};
    \end{tikzpicture}
    \\[0.75cm]
    Además, según Larson R. (2009) en su texto \emph{Cálculo Integral - Matemáticas 2} expresa que:
    \vspace{1ex}
    \begin{quote}
    ``La expresión $\displaystyle \int f(x)\,dx$ se lee como la antiderivada o primitiva de $f$ con respecto a $x$, el diferencial de $x$ sirve para identificar a $x$ como la variable de integración. El término \emph{integral indefinida} es sinónimo de antiderivada."
    \end{quote}
    \vspace{0.5cm}
    \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]
    {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{A lo largo de este texto, usted encontrará conforme vea las distintas técnicas y casos de integrandos particulares, la complejidad en el desarrollo de los mismos, como integrar cada función producirá una constante $C$, solo se asumirá en el resultado final escrito como la suma de todas las constantes de las integrales resueltas, de manera que $C = C_1 + C_2 + C_3 + \ldots + C_n$}
    \vspace{0.6cm}
    A continuación se presentará una lista de ejercicios con un orden aleatorio de dificultad y algunos ejemplos previamente explicados para ayudar a comprender el principio básico de la integración inmediata por medio del uso de la tabla.
    \\[0.6cm]
    \begin{tabular}{llllp{8cm}}
    \textbf{Ej. 1.1)} $ \displaystyle \int x + 3\,dx$ & = & $\displaystyle \int x\,dx + \int 3\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[5mm]
                                   & = & $\displaystyle \int x\,dx + 3 \int dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont en la integral de la derecha se extrajo el factor 3 fuera de la integral como una constante}     \\[1mm]
                   & = & $\displaystyle \frac{x^2}{2} + 3x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}
    \end{tabular}
    \newpage
    \begin{tabular}{llllp{3cm}}
    \textbf{Ej. 1.2)} $ \displaystyle \int \frac{x^2 + x + 1}{\sqrt{x}}\,dx$ & = & $\displaystyle \int \frac{x^2}{\sqrt{x}}\,dx + \int \frac{x}{\sqrt{x}}\,dx + \int \frac{1}{\sqrt{x}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]
                                   & = & $\displaystyle \int \frac{x^2}{x^{1/2}}\,dx + \int \frac{x}{x^{1/2}}\,dx + \int \frac{1}{x^{1/2}}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont escribir los radicales en forma de potencia}    \\[1mm]
                                    & = & $\displaystyle \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{- 1/2}\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]
                                   & = & $\displaystyle \frac{x^{5/2}}{5/2} + \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar} \\[6mm]
                                   & = & $\displaystyle \frac{2}{5}\,x^{5/2} + \frac{2}{3}\,x^{3/2} + 2\,\sqrt{x} + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}
    \end{tabular}
    \reversemarginpar
    \marginnote{
    \colorbox{yellow!30}{
    \begin{minipage}{2.5cm}
    \begin{spacing}{0.55}
    \fontsize{7}{14}\selectfont Las funciones irracionales (raíces) cuentan como funciones de potencia.
    \vspace{-8pt}
    \end{spacing}
    \end{minipage}}}
    \\[0.9cm]
    \begin{tabular}{llllp{4cm}}
    \textbf{Ej. 1.3)} $ \displaystyle \int (x + 1)(3x - 2)\,dx$ & = & $\displaystyle \int 3x^2 + x - 2\,dx$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont multiplicar factores y agrupar términos semejantes} \\[1mm]
                                  & = & $\displaystyle 3 \int x^2\,dx + \int x\,dx - 2 \int dx$ & $\longleftarrow$
    &  {\fontsize{8}{0}\selectfont separar en 3 integrales} \\[5mm]
                                  & = & $\displaystyle x^3 + \frac{x^2}{2} - 2x + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}
    \end{tabular}
    \\[0.9cm]
    \begin{tabular}{llllp{4cm}}
    \textbf{Ej. 1.4)} $ \displaystyle \int \sec y(\tan y - \sec y)\,dy$ & = & $\displaystyle \int \sec y \tan y - \sec^2y\,dy $ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont reescribir} \\[5mm]
                                   & $=$ & $\displaystyle \int \sec y \tan y\,dy - \int \sec^2 y\,dy$ &               $\longleftarrow$ & {\fontsize{8}{0}\selectfont separar en 2 integrales} \\[7mm]
                                   & $=$ & $\displaystyle \sec y - \tan y + C$ & $\longleftarrow$  & {\fontsize{8}{0}\selectfont integrar}
    \end{tabular}
    \\[0.9cm]
    \begin{tabular}{llllp{6cm}}
    \textbf{Ej. 1.5)} $ \displaystyle \int 2\pi y(8 - y^{3/2})\,dy$ & = & $\displaystyle 2\pi \int 8y - y^{5/2}\,dy$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont extraer el factor $2\pi$ fuera de la integral como una constante y reescribir la función}       \\[3mm]
                    & = & $\displaystyle 2\pi \left[4y^2 - \frac{y^{7/2}}{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont integrar}    \\[6mm]
                    & = & $\displaystyle 2\pi \left[4y^2 - \frac{2}{7}\, y^{7/2} \right] + C$ & $\longleftarrow$ & {\fontsize{8}{0}\selectfont simplificar}
    \end{tabular}
    \\
    \vfill
    \tcbsidebyside[sidebyside adapt=left, bicolor, colback=gris!90, colbacklower=gris!25, fonttitle=\bfseries, drop shadow, sidebyside gap=2mm]
    {\hspace*{-0.55cm} {\Huge {\bfseries\textcolor{white!100}{!}}} \hspace*{-0.15cm}}{El lector observará conforme vea los ejercicios elaborados de este texto que algunos de los pasos efectuados en los ejemplos 1.1 al 1.5 en la práctica son omitidos, esto ocurrirá a medida que se familiarice con las reglas básicas de integración.}
    \newpage
    Calcular las siguientes integrales
    \\[0.55cm]
    \hspace*{-0.25cm}
    %----------------------------------------------------------
    %                     LISTA DE EJERCICIOS
    %----------------------------------------------------------
    $
    {\setlength{\arraycolsep}{10pt}
    \begin{array}{*3{>{\displaystyle}l}}
    \textbf{1.6)} \int 2x - 3x^2\,dx & \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx & \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx \\[6mm]
    \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx  & \textbf{1.10)} \int \sqrt[3]{x^2}\,dx & \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx \\[6mm]
    \textbf{1.12)} \int (2t^2 - 1)^2\,dt & \textbf{1.13)} \int y^2\sqrt{y}\,dy & \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx  \\[6mm]
    \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt & \textbf{1.16)} \int \tan^2y + 1\,dy & \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx \\[6mm]
    \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx & \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx & \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx \\[6mm]
    \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx & \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx & \textbf{1.23)} \int \frac{x^2}{x^2 + 1}\,dx \\[6mm]
    \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx & \textbf{1.25)} \int \tan^2x\,dx & \textbf{1.26)} \int \cot^2x\,dx \\[6mm]
    \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}\,dx & \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx & \textbf{1.29)} \int (2^x + 3^x)^2\,dx \\[6mm]
    \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx & \textbf{1.31)} \int 3^xe^x\,dx & \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx \\[6mm]
    \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx & \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &
    \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy  \\[6mm]
    \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &   \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta & \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx \\[6mm]
    \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx & \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} & \textbf{1.41)} \int \frac{dx}{3x^2 + 5} \\[6mm]
    \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx & \textbf{1.43)} \int \left(y^2 - \frac{1}{y^2} \right)^3 dy & \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx \\[6mm]
    \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx & \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt  & \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}}
    \\[6mm]
    \textbf{1.48)} \int \frac{dx}{\sen x \cos x} & \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta & \textbf{1.50)} \int (\tan x + \sec x)^2\,dx  \\[6mm]
    \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx & \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx & \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi  \\[6mm]
    \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx & \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx  & \textbf{1.56)} \int 1^x\,dx
    \end{array}}
    $
    \newpage
    \hspace*{-0.35cm}
    $
    {\setlength{\arraycolsep}{10pt}
    \begin{array}{*3{>{\displaystyle}l}}
    \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx & \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx & \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx \\[6mm]
    \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx  & \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx & \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx \\[6mm]
    \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx & \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx & \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx
    \end{array}}
    $
    \\[1.5cm]
    \textbf{\huge Solución}
    \\
    \rule{21cm}{1ex}
    \\[1ex]

    %-------------------------------------------------------------
    %                           EJERCICIO 1.6
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.6)} \int 2x - 3x^2\,dx &= \int 2x\,dx - \int 3x^2\,dx = 2 \int x\,dx - 3 \int x^2\,dx = \cancel{2}\left(\frac{x^2}{\cancel{2}} \right) - \cancel{3} \left(\frac{x^3}{\cancel{3}} \right) + C \\[3mm]
                       &=  \fboxsep=5pt\colorbox{gris!40}{$x^2 - x^3 + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.7
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.7)} \int 4x^3 + 6x^2 - 1\,dx &= \int 4x^3\,dx + \int 6x^2\,dx - \int dx                                                                                           = 4 \int x^3\,dx + 6 \int x^2\,dx - \int dx
                                           \\[3mm]
                                           &= \cancel{4} \left(\frac{x^4}{\cancel{4}} \right) + 6 \left( \frac{x^3}{3} \right) - x + C  =  \fboxsep=5pt\colorbox{gris!40}{$x^4 + 2x^3 - x + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.8
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.8)} \int x^{3/2} + 2x + 1\,dx &= \int x^{3/2}\,dx + 2 \int x\,dx + \int dx
                                                                                    = \frac{x^{5/2}}{5/2} + \cancel{2} \left(\frac{x^2}{\cancel{2}} \right) + x + C
                                                                                    \\[3mm]
                                                            &=  \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{5}\,x^{5/2} + x^2 + x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.9
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.9)} \int \sqrt{x} + \frac{1}{2\sqrt{x}}\,dx &= \int \sqrt{x}\,dx + \int \frac{dx}{2\sqrt{x}} = \int x^{1/2}\,dx + \frac{1}{2} \int x^{-1/2}\,dx = \frac{x^{3/2}}{3/2} + \frac{1}{\cancel{2}} \left(\frac{x^{1/2}}{1/ \cancel{2}} \right) + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{3}\,x^{3/2} + x^{1/2} + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.10
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.10)} \int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{5}\,x^{5/3} + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.11
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.11)} \int \frac{x^2 + 2x - 3}{x^4}\,dx &= \int \frac{x^2}{x^4}\,dx + 2 \int \frac{x}{x^4}\,dx - 3 \int \frac{dx}{x^4} = \int x^{-2}\,dx + 2 \int x^{-3}\,dx - 3 \int x^{-4}\,dx              
    \end{align*}
    \begin{align*}
    &= \frac{x^{-1}}{- 1} + \cancel{2} \left(\frac{x^{-2}}{- \cancel{2}} \right) - \cancel{3} \left(\frac{x^{-3}}{- \cancel{3}} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.12
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.12)} \int (2t^2 - 1)^2\,dt &= \int 4t^4 - 4t^2 + 1\,dt = 4 \int t^4\,dt - 4 \int t^2\,dt + \int dt = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{4}{5} \, t^5 - \frac{4}{3} \, t^3 + t + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.13
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.13)} \int y^2\sqrt{y}\,dy = \int y^{5/2}\,dy = \frac{y^{7/2}}{7/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2}{7}\,y^{7/2} + C$}  
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.14
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.14)} \int 2 \sen x + 3 \cos x\,dx = 2 \int \sen x\,dx + 3 \int \cos x\,dx = \fboxsep=5pt\colorbox{gris!40}{$- 2 \cos x + 3 \sen x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.15
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.15)} \int \frac{1 - t^3 - t}{t^2 + 1}\,dt &= \int \frac{1 - t(t^2 + 1)}{t^2 + 1} = \int \frac{dt}{t^2 + 1} - \int \frac{t (\cancel{t^2 + 1})}{\cancel{t^2 + 1}}\,dt = \fboxsep=5pt\colorbox{gris!40}{$\arctan t - \displaystyle \frac{t^2}{2} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.16
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.16)} \int \tan^2y + 1\,dy = \int \sec^2y - 1 + 1\,dy = \fboxsep=5pt\colorbox{gris!40}{$\tan y + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.17
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.17)} \int x^2 + \frac{1}{(3x)^2}\,dx &= \int x^2\,dx + \int \frac{dx}{9x^2}                                          = \int x^2\,dx + \frac{1}{9} \int x^{-2}\,dx                                                                = \frac{x^3}{3} + \frac{1}{9}\left(\frac{x^{-1}}{-1}                                                                    \right) + C \\[3mm]
                               &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle                                                                        \frac{x^3}{3} - \frac{1}{9x} + C $}  
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.18
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.18)} \int \left(1 - \frac{1}{x^2} \right)\sqrt{x\sqrt{x}}\,dx &= \int \left(1 - \frac{1}{x^2} \right) \left(x^{3/2} \right)^{1/2}\,dx = \int \left(1 - \frac{1}{x^2} \right)x^{3/4}\,dx \\[3mm]
    &= \int x^{3/4}\,dx - \int x^{-5/4}\,dx = \frac{x^{7/4}}{7/4} - \left(- \frac{x^{-1/4}}{1/4} \right) + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{4}{7}\,x^{7/4} + \frac{4}{x^{1/4}} + C $}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.19
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.19)} \int \frac{{(\sqrt{2x} - \sqrt[3]{3x})}^2}{x}\,dx &= \int \frac{2x - 2\sqrt{2x}\sqrt[3]{3x} + {(\sqrt[3]{3x})}^2}{x}\,dx \\[3mm]
    &= 2 \int dx - 2 \int \frac{\sqrt{2}\,x^{1/2}\,\sqrt[3]{3}\,{x}^{1/3}}{x}\,dx + \int \frac{\sqrt[3]{9}\,x^{2/3}}{x}\,dx \\[3mm]
    &= 2 \int dx - 2 \sqrt{2} \sqrt[3]{3} \int x^{-1/6}\,dx + \sqrt[3]{9} \int x^{-1/3}\,dx \\[3mm]
    &= 2x - 2 \sqrt{2} \sqrt[3]{3} \left(\frac{x^{5/6}}{5/6} \right) + \sqrt[3]{9} \left( \frac{x^{2/3}}{2/3} \right) + C
    \end{align*}
    \begin{align*}
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle 2x - \frac{12}{5} \sqrt{2} \sqrt[3]{3}\,x^{5/6} + \frac{3}{2} \sqrt[3]{9}\,x^{2/3} + C $}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.20
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.20)} \int \sqrt[3]{x\sqrt{\frac{2}{x}}}\,dx &= \int \sqrt[3]{\sqrt{\frac{2x^2}{x}}}\,dx = \int \sqrt[3]{\sqrt{2x}}\,dx  = \int {\left[(2x)^{1/2} \right]}^{1/3}\,dx = \int (2x)^{1/6}\,dx  \\[3mm]
    &= \sqrt[6]{2} \left(\frac{x^{7/6}}{7/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt[6]{2}}{7}\,x^{7/6} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.21
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.21)} \int \frac{2^{x + 1} - 5^{x - 1}}{10^x}\,dx &= \int \frac{2^x\,2}{10^x}\,dx - \int \frac{5^x}{5\,10^x}\,dx = 2 \int \left(\frac{2}{10} \right)^x\,dx - \frac{1}{5} \int \left(\frac{5}{10} \right)^x\,dx \\[3mm]
    &= 2 \left[ \frac{(1/5)^x}{\ln(1/5)} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln(1/2)} \right] + C = 2 \left[ \frac{(1/5)^x}{\ln 1 - \ln 5} \right] - \frac{1}{5} \left[ \frac{(1/2)^x}{\ln 1- \ln 2} \right] + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{5 \ln 2} \left(\frac{1}{2} \right)^x - \frac{2}{\ln 5} \left(\frac{1}{5} \right)^x + C $}
    \end{align*}
    %\reversemarginpar
    \marginnote{
    \colorbox{yellow!30}{
    \begin{minipage}{2.5cm}
    \begin{spacing}{0.55}
    \fontsize{7}{14}\selectfont Es menester destacar que en el ejercicio 1.23 no era necesario aplicar el algoritmo de la división, simplemente con sumar y restar el factor 1 en el numerador y separar las fracciones se obtendría el mismo resultado, este artificio de sumar y restar, multiplicar y dividir elementos será de gran utilidad para la resolución de un gran número de ejercicios presentados en este texto.
    \vspace{-8pt}
    \end{spacing}
    \end{minipage}}
    }

    %-------------------------------------------------------------
    %                          EJERCICIO 1.22
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.22)} \int \frac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\,dx &= \int \frac{\sqrt{x^4 + \displaystyle \frac{1}{x^4} + 2}}{x^3}\,dx = \int \frac{\displaystyle \sqrt{\frac{x^8 + 2x^4 + 1}{x^4}}}{x^3}\,dx \\[3mm]
    &= \int \frac{\sqrt{x^8 + 2x^4 + 1}}{x^5}\,dx = \int \frac{\sqrt{(x^4 + 1)^2}}{x^5}\,dx = \int \frac{x^4 + 1}{x^5}\,dx  \\[3mm]
    &= \int \frac{dx}{x} + \int x^{-5}\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \ln \big| x \big| - \frac{1}{4x^4} + C$}
    \end{align*}
    \vspace{1cm}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.23
    %-------------------------------------------------------------


    \textbf{1.23)} $ \displaystyle \int \frac{x^2}{x^2 + 1}\,dx$ \quad al aplicar la división de polinomios
    \quad
    $
    \begin{array}{cccc|ccc}
       & \cancel{x^2} & + & 0 & x^2 & + & 1 \\
    \cline{5-7}
     - & \cancel{x^2} & - & 1 & 1   &   &   \\
    \cline{2-4}  
       &              & - & 1 &     &   &
    \end{array}
    $
    \\[0.25cm]
    \begin{align*}
    \mbox{La integral se convierte en} \int \frac{x^2}{x^2 + 1}\,dx &= \int \frac{(x^2 + 1)1 - 1}{x^2 + 1}\,dx = \int \frac{\cancel{x^2 + 1}}{\cancel{x^2 + 1}}\,dx - \int \frac{dx}{x^2 + 1}       \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle x - \arctan x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.24
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.24)} \int \frac{e^{3x} + 1}{e^x + 1}\,dx &= \int \frac{\cancel{(e^x + 1)}(e^{2x} - e^x + 1)}{\cancel{e^x + 1}}\,dx = \int e^{2x} - e^x + 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle\frac{e^{2x}}{2} - e^x + x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.25
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.25)} \int \tan^2x\,dx = \int \sec^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$\tan x - x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.26
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.26)} \int \cot^2x\,dx = \int \csc^2x - 1\,dx = \fboxsep=5pt\colorbox{gris!40}{$- \cot x - x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.27
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.27)} \int \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx &= \int \frac{\sqrt{1 + x^2}}{\sqrt{1 - x^4}}\,dx + \int \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^4}}\,dx \\[3mm]
    &= \int \sqrt{\frac{\cancel{1 + x^2}}{(1 - x^2)\cancel{(1 + x^2)}}}\,dx + \int \sqrt{\frac{\cancel{1 - x^2}}{\cancel{(1 - x^2)}(1 + x^2)}}\,dx \\[3mm]
    &= \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{dx}{\sqrt{x^2 + 1}} \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\arcsen x + \ln(x + \sqrt{x^2 + 1}) + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.28
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.28)} \int \frac{(1 - x)^3}{x\sqrt[3]{x}}\,dx &= \int \frac{1 - 3x^2 + 3x - x^3}{x\,x^{1/3}}\,dx
    \\[3mm]
    &= \int x^{-4/3}\,dx - 3 \int x^{2/3}\,dx + 3 \int x^{-1/3}\,dx - \int x^{5/3}\,dx \\[3mm]
    &= \frac{x^{-1/3}}{-1/3} - 3 \, \frac{x^{5/3}}{5/3} + 3 \, \frac{x^{2/3}}{2/3} - \frac{x^{8/3}}{8/3} + C
    \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{3}{x^{1/3}} - \frac{9}{5}\,x^{5/3} + \frac{9}{2}\,x^{2/3} - \frac{3}{8}\,x^{8/3} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.29
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.29)} \int (2^x + 3^x)^2\,dx &= \int {(2^x)}^2 + 2(2^x)(3^x) + {(3^x)}^2\,dx = \int 4^x\,dx + 2 \int 6^x\,dx + \int 9^x\,dx \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{\ln 4}\,4^x + \frac{2}{\ln 6}\,6^x + \frac{1}{\ln 9}\,9^x + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.30
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.30)} \int (nx)^{\frac{1 - n}{n}}\,dx &= \int n^{\frac{1 - n}{n}}\,x^{\frac{1 - n}{n}}\,dx
    = n^{\frac{1 - n}{n}} \int x^{\frac{1 - n}{n}}\,dx = n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - n}{n} + 1}}{\frac{1 - n}{n} + 1} \right) + C \\[2mm]
    &= n^{\frac{1 - n}{n}} \left(\displaystyle \frac{x^{\frac{1 - \cancel{n} + \cancel{n}}{n}}}{\frac{1 - \cancel{n} + \cancel{n}}{n}} \right) + C = n^{\frac{1 - n}{n}}\,n\,x^{1/n} + C = n^{1/n}\,x^{1/n} + C \\[2mm]
    &= (nx)^{1/n} + C = \fboxsep=5pt\colorbox{gris!40}{$\sqrt[n]{nx} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.31
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.31)} \int 3^xe^x\,dx &= \int (3e)^x\,dx = \frac{(3e)^x}{\ln(3e)} + C = \frac{(3e)^x}{\ln 3 + \ln e} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3^xe^x}{\ln 3 + 1} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.32
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.32)} \int {\left(a^{2/3} - x^{2/3} \right)}^3\,dx &= \int {(a^{2/3})}^3 - 3{(a^{2/3})}^2x^{2/3} + 3a^{2/3}{(x^{2/3})}^2 - {(x^{2/3})}^3\,dx \\[3mm]
    &= \int a^2 - 3a^{4/3}x^{2/3} + 3a^{2/3}x^{4/3} - x^2\,dx \\[3mm]
    &= a^2 \int dx - 3a^{4/3} \int x^{2/3}\,dx + 3a^{2/3} \int x^{4/3}\,dx - \int x^2\,dx \\[3mm]
    &= a^2x - 3a^{4/3}\left(\frac{x^{5/3}}{5/3} \right) + 3a^{2/3} \left(\frac{x^{7/3}}{7/3} \right) - \frac{x^3}{3} + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle a^2x - \frac{9a^{4/3}}{5}\,x^{5/3} + \frac{9a^{2/3}}{7}\,x^{7/3} - \frac{x^3}{3} + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.33
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.33)} \int \frac{\left(x^m - x^n \right)^2}{\sqrt{x}}\,dx &= \int \frac{x^{2m} - 2x^mx^n + x^{2n}}{\sqrt{x}}\,dx  \\[3mm]
    &= \int x^{2m - 1/2}\,dx - 2 \int x^{m + n - 1/2}\,dx + \int x^{2n - 1/2}\,dx \\[3mm]
    &= \frac{x^{2m - 1/2 + 1}}{2m - 1/2 + 1} - 2 \left(\frac{x^{m + n - 1/2 + 1}}{m + n - 1/2 + 1} \right) + \frac{x^{2n - 1/2 + 1}}{2n - 1/2 + 1} + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2 \left( \frac{x^{\frac{4m + 1}{2}}}{4m + 1} \right) - 4 \left( \frac{x^{\frac{2m + 2n + 1}{2}}}{2m + 2n + 1} \right) + 2 \left( \frac{x^{\frac{4n + 1}{2}}}{4n + 1} \right) + C$}
    \end{align*}

    %-------------------------------------------------------------
    %                          EJERCICIO 1.34
    %-------------------------------------------------------------

    \begin{align*}
    \textbf{1.34)} \int \frac{\left(a^x - b^x \right)^2}{a^x b^x}\,dx &= \int \frac{a^{2x} - 2a^xb^x + b^{2x}}{a^xb^x}\,dx = \int \frac{a^{2x}}{a^xb^x}\,dx -2 \int \frac{\cancel{a^xb^x}}{\cancel{a^xb^x}}\,dx + \int \frac{b^{2x}}{a^xb^x}\,dx \\[2mm]
    &= \int \left(\frac{a}{b} \right)^xdx - 2 \int dx + \int \left(\frac{b}{a} \right)^xdx \\[2mm]
    &= \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln (a/b)} - 2x + \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln (b/a)} + C = \frac{\displaystyle \left(\frac{a}{b} \right)^x}{\ln a - \ln b} - 2x - \frac{\displaystyle \left(\frac{b}{a} \right)^x}{\ln a - \ln b} + C \\[2mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{1}{\ln a - \ln b} \left[ \left(\frac{a}{b} \right)^x - \left(\frac{b}{a} \right)^x \right] - 2x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.35
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.35)} \int 4y^3 + \frac{2}{y^3}\,dy &= 4 \int y^3\,dy + 2 \int \frac{dy}{y^3} = \cancel{4}\left(\frac{y^4}{\cancel{4}} \right) - \cancel{2}\left(\frac{1}{- \cancel{2}y^2} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle y^4 + \frac{1}{y^2} + C $}
    \end{align*}
    \vspace{-0.65cm}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.36
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.36)} \int \left(\frac{1}{\sqrt{2}\sen x} - 1\right)^2\,dx &= \int \left(\frac{1}{\sqrt{2}\sen x} \right)^2 - 2\left(\frac{1}{\sqrt{2} \sen x} \right) + 1\,dx \\[3mm]
    &= \int \frac{dx}{2 \sen^2x} - 2 \int \frac{dx}{\sqrt{2} \sen x} + \int dx = \frac{1}{2} \int \csc^2x\,dx - \sqrt{2} \int \csc x\,dx  + \int dx \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{1}{2} \cot x - \sqrt{2} \ln \big|\csc x - \cot x \big| + x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.37
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.37)} \int \frac{\sen \theta + \sen \theta\,\tan^2\theta}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta(1 + \tan^2\theta)}{\sec^2\theta}\,d\theta = \int \frac{\sen \theta\,\cancel{\sec^2\theta}}{\cancel{\sec^2\theta}}\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \cos \theta + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.38
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.38)} \int \frac{\sen x}{1 - \sen^2x}\,dx &= \int \frac{\sen x}{\cos^2x}\,dx = \int \frac{1}{\cos x}\,\frac{\sen x}{\cos x}\,dx = \int \sec \theta\,\tan \theta\,d\theta = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \sec \theta + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.39
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.39)} \int \frac{\sen(2x)}{\sen x}\,dx &= \int \frac{2\,\cancel{\sen x}\,\cos x}{\cancel{\sen x}}\,dx = \fboxsep=5pt\colorbox{gris!40}{$2\,\sen x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.40
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.40)} \int \frac{dx}{(a + b) - (a - b)^2} &= \int \frac{dx}{(a - b) \left[\displaystyle \frac{a + b}{a - b} - x^2 \right]} = \frac{1}{a - b} \int \frac{dx}{\left(\sqrt{\frac{a + b}{a - b}} \right)^2 - x^2} \\[3mm]
    &= \frac{1}{2(a - b)\displaystyle \sqrt{\frac{a + b}{a - b}}} \ln \left|\displaystyle \frac{\displaystyle \sqrt{\frac{a + b}{a - b}} + x}{\displaystyle \sqrt{\frac{a + b}{a - b}} - x} \right| + C \\[3mm]
    &= \frac{1}{2\sqrt{(a + b)(a - b)}} \ln \left| \frac{\displaystyle  \frac{\sqrt{a + b} \ + \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}}{\displaystyle  \frac{\sqrt{a + b} \ - \ \sqrt{a - b}x}{\cancel{\sqrt{a - b}}}} \right| + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{1}{2 \sqrt{a^2 - b^2}} \ln \left| \frac{\sqrt{a + b} + \sqrt{a - b}x}{\sqrt{a + b} - \sqrt{a - b}x} \right| + C$}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.41
    %-------------------------------------------------------------



    \begin{align*}
    \textbf{1.41)} \int \frac{dx}{3x^2 + 5} &= \int \frac{dx}{3\left(x^2 + \frac{5}{3} \right)}
    = \frac{1}{3} \int \frac{dx}{x^2 + \left(\sqrt{\frac{5}{3}} \right)^2} = \frac{1}{3} \left[\frac{1}{\sqrt{\frac{5}{3}}} \arctan \left(\frac{x}{\sqrt{\frac{5}{3}}} \right) \right] + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{\sqrt{3}}{3\sqrt{5}} \arctan \left(\frac{\sqrt{3} x}{\sqrt{5}} \right) + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.42
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.42)} \int \left(\sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx &= \int {(\sqrt{x})}^2 - 2 \left(\cancel{\sqrt{x}} \frac{1}{\cancel{\sqrt{x}}} \right) + \left(\frac{1}{\sqrt{x}} \right)^2 dx = \int x^2\,dx - 2 \int dx + \int \frac{dx}{x} \\[3mm]
    & = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{x^3}{3} - 2x + \ln \big| x \big| + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.43
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.43)}\int \left(y^2 - \frac{1}{y^2} \right)^3 dy &= \int {(y^2)}^3 - 3{(y^2)}^2\left(\frac{1}{y^2} \right) + 3y^2 \left(\frac{1}{y^2} \right)^2 - \left(\frac{1}{y^2} \right)^{\!\! 3} dy \\[3mm]
    &= \int y^6\,dy - 3 \int y^2\,dy + 3 \int y^{-2}\,dy - \int y^{-6}\,dy \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{y^7}{7} - y^3 - \frac{3}{y} + \frac{1}{5y^5} + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.44
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.44)} \int \left(e^{x/a} - e^{- x/a} \right)^2\,dx &= \int \left( e^{x/a} \right)^2 - 2 e^{x/a} e^{- x/a} + \left( e^{- x/a} \right)^2 dx \\[3mm]
    &= \int e^{2x/a} dx - 2 \int e^0\,dx + \int e^{-2x/a} dx \\[3mm]
    &= \frac{e^{2x/a}}{\frac{2}{a}} - 2x - \frac{e^{- 2x/a}}{\frac{2}{a}} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{a}{2} \left[e^{2x/a} - e^{- 2x/a} \right] - 2x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.45
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.45)} \int \frac{\sqrt{5x}}{5} + \frac{5}{\sqrt{5x}}\,dx &= \frac{\sqrt{5}}{5} \int x^{1/2}\,dx + \sqrt{5} \int x^{-1/2}\,dx = \frac{\sqrt{5}}{5} \left( \frac{2}{3} x^{3/2} \right) + \sqrt{5}(2 \sqrt{x}) + C \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{2\sqrt{5}}{15}\,x^{3/2} + 2 \sqrt{5x} + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.46
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.46)} \int \frac{4}{\sqrt{e^t}}\,dt = 4 \int {(e^t)}^{-1/2} = - 4 \, \frac{e^{-t/2}}{1/2} + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle - \frac{8}{\sqrt{e^t}} + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.47
    %-------------------------------------------------------------


    %\vspace*{-1cm}
    \begin{align*}
    \textbf{1.47)} \int \frac{dx}{\sqrt{7 - 5x^2}} &= \int \frac{dx}{\sqrt{5\left(\frac{7}{5} - x^2 \right)}} = \frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(\sqrt{\frac{7}{5}} \right)^2 - x^2}} = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{\sqrt{5}}{5} \arcsen \left(\frac{\sqrt{5}x}{\sqrt{7}} \right) + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.48
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.48)} \int \frac{dx}{\sen x \cos x} &= \int \frac{\sen^2x + \cos^2x}{\sen x \cos x}\,dx = \int \frac{\sen^2x}{\sen x \cos x}\,dx + \int \frac{\cos^2x}{\sen x \cos x}\,dx = \int \tan x\,dx + \\[3mm]
    & \int \cot x\,dx = - \ln \big| \cos x \big| + \ln \big| \sen x \big| + C = \ln \left| \frac{\sen x}{\cos x} \right| + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \ln \big| \tan x \big | + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.49
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.49)} \int \frac{1 - \cos^2\theta}{\cos^2\theta}\,d \theta = \int \frac{d \theta}{\cos^2\theta} + \int \frac{\cancel{\cos^2 \theta}}{\cancel{\cos^2 \theta}}\,d \theta = \int \sec^2 \theta\,d \theta + \int d \theta
    = \fboxsep=5pt\colorbox{gris!40}{$ \tan \theta + \theta + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.50
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.50)} \int (\tan x + \sec x)^2\,dx &= \int \tan^2x + 2 \tan x\,\sec x + \sec^2x\,dx
    \\[3mm]
    &= \int \sec^2x - 1 + 2 \tan x\,\sec x + \sec^2x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2\tan x + 2\sec x - x + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.51
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.51)} \int \sqrt{x \sqrt{x}}\,dx = \int \left(x^{3/2} \right)^{1/2}dx = \int x^{3/4}dx = \frac{x^{7/4}}{7/4} + C = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{4}{7}\,x^{7/4} + C$}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.52
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.52)} \int \frac{9x^6 - 4}{3x^3 + 2}\,dx &= \int \frac{(3x^3 - 2) \cancel{(3x^3 + 2)}}{\cancel{(3x^3 + 2)}}\,dx = \int 3x^3\,dx - \int 2\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{3}{4}\,x^4 - 2x + C$}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.53
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.53)} \int \frac{\csc \phi}{\csc \phi - \sen \phi}\,d\phi &= \int \frac{\displaystyle \frac{1}{\sen \phi}}{\displaystyle \frac{1}{\sen \phi} - \sen \phi}\,d\phi = \int \frac{\displaystyle \frac{1}{\cancel{\sen \phi}}}{\displaystyle \frac{1 - \sen^2\phi}{\cancel{\sen \phi}}}\,d \phi = \int \frac{1}{\cos^2\phi}\,d\phi = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \tan \phi + C$}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.54
    %-------------------------------------------------------------


    \marginnote{
    \colorbox{yellow!30}{
    \begin{minipage}{2.5cm}
    \begin{spacing}{0.55}
    \fontsize{7}{14}\selectfont Recuerde que la expresión $\int \frac{\sen x}{\cos^2x}\,dx$ ya había aparecido antes en el ejercicio 1.38 por lo que se omitieron algunos detalles de solución.
    \vspace{-8pt}
    \end{spacing}
    \end{minipage}}}
    \vspace*{-0.25cm}
    \begin{align*}
    \textbf{1.54)} \int \frac{\sen x + \tan x}{\cos x}\,dx &= \int \frac{\sen x}{\cos x}\,dx + \int \frac{\tan x}{\cos x}\,dx = \int \tan x\,dx + \int \frac{\sen x}{\cos^2x}\,dx \\[3mm]
    &= \int \tan x\,dx + \int \sec x \tan x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle - \ln \big| \cos x \big| + \sec x + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.55
    %-------------------------------------------------------------

    \marginnote{
    \colorbox{yellow!30}{
    \begin{minipage}{2.5cm}
    \begin{spacing}{0.55}
    \fontsize{7}{14}\selectfont La integral del ejercicio 1.56 es de la forma $a^x$ pero $a = 1$ por lo que no se puede aplicar la forma de la tabla, pero si las propiedades de los logaritmos.
    \vspace{-8pt}
    \end{spacing}
    \end{minipage}}}


    \begin{align*}
    \textbf{1.55)} \int \frac{\sqrt{x^2 - x} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx &= \frac{\sqrt{x(x - 1)} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx \\[3mm]
    &= \int \frac{\sqrt{x}\sqrt{x - 1} - e^x\sqrt{x - 1}}{\sqrt{x - 1}}\,dx = \int \frac{\cancel{\sqrt{x - 1}}(\sqrt{x} - e^x)}{\cancel{\sqrt{x - 1}}}\,dx \\[3mm]
    &= \int \sqrt{x}\,dx - \int e^x\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{2}{3}\,x^{3/2} - e^x + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.56
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.56)} \int 1^x\,dx = \int e^{\ln 1^x}dx = \int e^{x\,\ln 1}dx = \int e^0dx = \int dx = \fboxsep=5pt\colorbox{gris!40}{$ x + C$}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.57
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.57)} \int \sqrt[3]{x^5}\,x^{-4/3}(x^3 - 1)\,dx &= \int x^{5/3}x^{-4/3}(x^3 - 1)\,dx = \int x^{1/3}(x^3 - 1)\,dx  \\[3mm]
    &= \int x^{10/3}dx - \int x^{1/3}dx = \frac{x^{13/3}}{13/3} + \frac{x^{4/3}}{4/3} + C
    \end{align*}
    \begin{align*}
    = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{3}{13}\,x^{13/3} + \frac{3}{4}\,x^{4/3} + C $}
    \end{align*}



    %-------------------------------------------------------------
    %                          EJERCICIO 1.58
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.58)} \int \frac{x - 1}{\sqrt{2x} - \sqrt{x}}\,dx &= \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{(\sqrt{2x} - \sqrt{x})(\sqrt{2x} + \sqrt{x})}\,dx = \int \frac{(x - 1)(\sqrt{2x} + \sqrt{x})}{2x - x}\,dx
    \\[3mm]
    &= \int \frac{\sqrt{2}x^{3/2} + x^{3/2} - \sqrt{2}x^{1/2} - x^{1/2}}{x}\,dx \\[3mm]
    &= \sqrt{2} \int x^{1/2}\,dx + \int x^{1/2}\,dx - \sqrt{2} \int x^{-1/2}\,dx - \int x^{- 1/2}\,dx \\[3mm]
    &= ( \sqrt{2} + 1) \int x^{1/2}\,dx - (\sqrt{2} + 1) \int x^{-1/2}\,dx \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$\displaystyle (\sqrt{2} + 1) \left[\frac{2}{3}\,x^{3/2} - 2x^{1/2} \right] + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.59
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.59)} \int \frac{\sqrt{5x^3}}{\sqrt[3]{3x}}\,dx &= \frac{\sqrt{5}}{\sqrt[3]{3}} \int \frac{x^{3/2}}{x^{1/3}}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \int x^{7/6}\,dx = \frac{\sqrt{5}}{\sqrt[3]{3}} \left( \frac{x^{13/6}}{13/6} \right) + C = \fboxsep=5pt\colorbox{gris!40}{$\displaystyle \frac{6 \sqrt{5}}{13 \sqrt[3]{3}}\,x^{13/6} + C$}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.60
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.60)} \int \frac{\tan x - \sen^2x + 4 \cos x}{3 \sen x}\,dx &= \int \frac{\tan x}{3 \sen x}\,dx - \int \frac{\sen^2x}{3 \sen x}\,dx + \int \frac{4 \cos x}{3 \sen x}\,dx \\[3mm]
    &= \frac{1}{3} \int \frac{\cancel{\sen x}}{\cos x \, \cancel{\sen x}}\,dx - \frac{1}{3} \int \sen x\,dx + \frac{4}{3} \int \cot x\,dx \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \frac{1}{3}\,\ln \big| \sec x + \tan x \big| + \frac{1}{3} \cos x + \frac{4}{3} \ln \big| \sen x \big| + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.61
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.61)} \int \left(\frac{1}{x} - x \right)^3\,dx &= \int \left(\frac{1}{x} \right)^3 - 3 \left(\frac{1}{x} \right)^2 x + 3 \left(\frac{1}{x} \right)\,x^2 - x^3\,dx \\[3mm]
    &= \int x^{-3}\,dx - 3 \int \frac{dx}{x} + 3 \int x\,dx - \int x^3\,dx \\[3mm]
    &= \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle - \frac{1}{2\,x^2} - 3\,\ln \big| x \big| + \frac{3}{2}\,x^2 - \frac{x^4}{4} + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.62
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.62)} \int \frac{e^{x + 2}}{e^{x + 1}}\,dx = \int \frac{\cancel{e^x}\,e^2}{\cancel{e^x}\,e}\,dx = \int e\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle ex + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.63
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.63)} \int x^{-2}(8x^5 - 6x^4 - x^{-1})\,dx = \int 8x^3 - 6x^2 - x^{-3}\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 2x^4 - 2x^3 + \frac{1}{2x^2} + C $}
    \end{align*}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.64
    %-------------------------------------------------------------


    \begin{align*}
    \textbf{1.64)} \int \frac{\ln (x^4)}{\ln x}\,dx = \int \frac{4 \, \cancel{\ln x}}{\cancel{\ln x}}\,dx = 4 \int dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle 4x + C $}
    \end{align*}
    \marginnote{
    \colorbox{yellow!30}{
    \begin{minipage}{2.5cm}
    \begin{spacing}{0.55}
    \fontsize{7}{14}\selectfont Observe con cuidado que en el ejercicio 1.64, la función logaritmo del numerador \emph{no está elevada a la cuarta potencia}, solo su argumento, por eso fue posible la aplicación de la propiedad y la posterior la simplificación de los términos.
    \vspace{-8pt}
    \end{spacing}
    \end{minipage}}}


    %-------------------------------------------------------------
    %                          EJERCICIO 1.65
    %-------------------------------------------------------------


    \vspace*{-0.8cm}
    \begin{align*}
    \textbf{1.65)} \int \frac{dx}{1 + \sen x}\,dx &= \int \frac{1 - \sen x}{(1 + \sen x)(1 - \sen x)}\,dx = \int \frac{1 - \sen x}{1 - \sen^2x}\,dx
    = \int \frac{1 - \sen x}{\cos^2x}\,dx
    \\[3mm]
    &= \int \sec^2x\,dx - \int \frac{\sen x}{\cos^2x}\,dx = \fboxsep=5pt\colorbox{gris!40}{$ \displaystyle \tan x - \sec x + C $}
    \end{align*}
    \end{document}

EDIT: Package fontenc with opion T1 is missing as well.
Add \usepackage[T1]{fontenc} to your preamble.

LaTeX.org

Text Formatting ⇒ Undefined old font command

Undefined old font command

Undefined old font command

Undefined old font command

Undefined old font command

Undefined old font command

Undefined old font command

Re: Undefined old font command

Undefined old font command

Re: Undefined old font command

Undefined old font command