100 Integrals - Rewritten
Forfatter:
josiah
Sidst opdateret:
4 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
just 100 integrals
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{bigints}
\DeclareMathOperator{\arcsec}{arcsec}
\DeclareMathOperator{\arccot}{arccot}
\DeclareMathOperator{\arccsc}{arccsc}
\addtolength{\jot}{0.75em}
\title{100 Integrals}
\author{Josiah Rondaris}
\begin{document}
\maketitle
\section{Basic Integrals to Remember}
\begin{align*}
&\int x^ndx=\frac{1}{n+1}x^{n+1}+C \\
&\int\sin\left(x\right)dx=-\cos\left(x\right)+C \\
&\int\cos\left(x\right)dx=\sin\left(x\right)+C \\
&\int\tan\left(x\right)dx=-\ln\left|\cos\left(x\right)\right|+C\mbox{ or }\ln\left|\sec\left(x\right)\right|+C \\
&\int\csc\left(x\right)dx=-\ln\left|\csc\left(x\right)+\cot\left(x\right)\right|+C \\
&\int\sec\left(x\right)dx=\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|+C \\
&\int\cot\left(x\right)dx=\ln\left|\sin\left(x\right)\right|+C\mbox{ or }-\ln\left|\csc\left(x\right)\right|+C \\
&\int a^xdx=\frac{1}{\ln\left(a\right)}a^x+C \\
&\int e^xdx=e^x+C \\
&\int\log_a\left(x\right)dx=x\log_a\left(x\right)-\frac{1}{\ln\left(a\right)}x+C \\
&\int\ln\left(x\right)dx=x\ln\left(x\right)-x+C
\end{align*}
\section{Advanced Integrals to Remember}
\begin{align*}
&\int\sec^3(x)dx=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C \\
&\int\arcsin(x)dx=x\arcsin(x)+\sqrt{1-x^2}+C \\
&\int\arccos(x)dx=x\arccos(x)-\sqrt{1-x^2}+C \\
&\int\arctan(x)dx=x\arctan(x)-\frac{1}{2}\ln(x^2+1)+C \\
&\int\sqrt{x^2+a^2}dx=\frac{1}{2}x\sqrt{x^2+a^2}+\frac{a^2}{2}\ln\left|\sqrt{x^2+a^2}+x\right|+C \\
&\int\sqrt{x^2-a^2}dx=\frac{1}{2}x\sqrt{x^2-a^2}-\frac{a^2}{2}\ln\left|\sqrt{x^2-a^2}+x\right|+C \\
&\int\sqrt{a^2-x^2}dx=\frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}x\sqrt{a^2-x^2}+C \\
&\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C \\
&\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C \\
&\int\frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C \\
&\int\frac{1}{\sqrt{x^2+a^2}}dx=\ln\left|x+\sqrt{x^2+a^2}\right|+C \\
&\int\frac{1}{\sqrt{x^2-a^2}}dx=\ln\left|x+\sqrt{x^2-a^2}\right|+C \\
&\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin\left(\frac{x}{a}\right)+C \\
&\int\sin^n(x)dx=-\frac{1}{n}\sin^{n-1}(x)\cos(x)+\frac{n-1}{n}\int\sin^{n-2}(x)dx \\
&\int\cos^n(x)dx=\frac{1}{n}\cos^{n-1}(x)\sin(x)+\frac{n-1}{n}\int\cos^{n-2}(x)dx
\end{align*}
\newpage
\section{The 100 Integrals}
\subsection*{Problem 1}
$$\int\frac{1}{\sqrt{x}\left(1+x\right)}dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
du&=\frac{1}{2\sqrt{x}}dx
\end{align*}
yields:
$$=2\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=2\arctan(u)+C$$
Undoing the substitution(s):
$$=2\arctan\left(\sqrt{x}\right)+C$$
\subsection*{Problem 2}
$$\int\frac{\sec^2\left(x\right)}{1+\tan\left(x\right)}dx$$
Substituting
\begin{align*}
u&=1+\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u}du$$
This integral is standard:
$$=\ln|u|+C$$
Undoing the substitution(s):
$$=\ln\left|1+\tan(x)\right|+C$$
\subsection*{Problem 3}
$$\int\sin\left(x\right)\sec\left(x\right)dx$$
Rewriting the integral:
$$=\int\frac{\sin(x)}{\cos(x)}dx$$
Substituting
\begin{align*}
u&=\cos(x) \\
du&=-\sin(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\int\ln|u|+C$$
Undoing the substitution(s):
$$=-\ln\left|\cos(x)\right|+C$$
\subsection*{Problem 4}
$$\int\frac{\csc\left(x\right)\cot\left(x\right)}{1+\csc^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\csc(x) \\
du&=-\csc(x)\cot(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=-\arctan(u)+C$$
Undoing the substitution(s):
$$=-\arctan(\csc(x))+C$$
\subsection*{Problem 5}
$$\int\frac{\tan\left(x\right)}{\cos^2\left(x\right)}dx$$
Rewriting the integral:
$$=\int\tan(x)\sec^2(x)dx$$
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
Yields:
$$=\int udu$$
This integral is standard:
$$=\frac{1}{2}u^2+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\tan^2(x)+C$$
\subsection*{Problem 6}
$$\int\csc^4\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\csc^2(x)\cdot\csc^2(x)dx \\
&=\int\csc^2(x)(\cot^2(x)+1)dx
\end{align*}
Substituting
\begin{align*}
u&=\cot(x) \\
du&=-\csc^2(x)dx
\end{align*}
Yields:
\begin{align*}
&=-\int(u^2+1)du \\
&=-\int u^2du-\int du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{3}u^3-u+C$$
Undoing the substitution(s):
$$=-\frac{1}{3}\cot^3(x)-\cot(x)+C$$
\subsection*{Problem 7}
$$\int x\tan^2\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int x(\sec^2(x)-1)dx \\
&=\int x\sec^2(x)dx- \int xdx
\end{align*}
Using integration by parts where
\begin{align*}
u&=x &v&=\tan(x) \\
du&=dx &dv&=\sec^2(x)dx
\end{align*}
yields:
$$=x\tan(x)-\int\tan(x)dx-\int xdx$$
Both integrals are standard:
$$=x\tan(x)+\ln|\cos(x)|-\frac{1}{2}x^2+C$$
\subsection*{Problem 8}
$$\int x^2\cos^2\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int x^2\left(\frac{1+\cos(2x)}{2}\right)dx \\
&=\frac{1}{2}\int x^2(1+\cos(2x))dx \\
&=\frac{1}{2}\int x^2dx+\frac{1}{2}\int x^2\cos(2x)dx
\end{align*}
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $x^2$ & $\cos(2x)$ \\
$-$ & $2x$ & $\frac{1}{2}\sin(2x)$ \\
$+$ & $2$ & $-\frac{1}{4}\cos(2x)$ \\
$-$ & $0$ & $-\frac{1}{8}\sin(2x)$ \\
\hline
\end{tabular}
\end{center}
yields:
\begin{align*}
&=\frac{1}{6}x^3+\frac{1}{2}\left(\frac{1}{2}x^2\sin(2x)+\frac{1}{2}x\cos(2x)-\frac{1}{4}\sin(2x)\right)+C \\
&=\frac{1}{6}x^3+\frac{1}{4}x^2\sin(2x)+\frac{1}{4}x\cos(2x)-\frac{1}{8}\sin(2x)+C
\end{align*}
\subsection*{Problem 9}
$$\int x^5\sqrt{2-x^3}dx$$
Rewriting the integral:
$$=\int x^2\cdot x^3\sqrt{2-x^3}dx$$
Substituting
\begin{align*}
u&=2-x^3 \\
2-u&=x^3 \\
-du&=3x^2dx
\end{align*}
yields:
\begin{align*}
&=-\frac{1}{3}\int(2-u)\sqrt{u}du \\
&=-\frac{1}{3}\int\left(2\sqrt{u}-u^{3/2}\right)du \\
&=-\frac{2}{3}\int\sqrt{u}du+\frac{1}{3}\int u^{3/2}du
\end{align*}
Both integrals are standard:
$$=-\frac{4}{9}u^{3/2}+\frac{2}{15}u^{5/2}+C$$
Undoing the substitution(s):
$$=-\frac{4}{9}(2-x^3)^{3/2}+\frac{2}{15}(2-x^3)^{5/2}+C$$
\subsection*{Problem 10}
$$\int\frac{1}{\sqrt{x^2+4}}dx$$
This integral is standard:
$$=\ln\left|x+\sqrt{x^2+4}\right|+C$$
\subsection*{Problem 11}
$$\int\frac{x^2}{\sqrt{25+x^2}}dx$$
Rewriting the integral:
\begin{align*}
=\int\frac{x^2}{\sqrt{25\big(1+\left(\frac{x}{5}\right)^2\big)}}dx \\
=\frac{1}{5}\int\frac{x^2}{\sqrt{1+(\frac{x}{5})^2}}dx
\end{align*}
Substituting
\begin{align*}
\tan\theta&=\frac{x}{5} \\
5\tan\theta&=x \\
5\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(5\tan\theta)^2}{\sqrt{1+\tan^2\theta}}\sec^2\theta d\theta \\
&=25\int\frac{\tan^2\theta\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta \\
&=25\int(\sec^2\theta-1)\sec\theta d\theta \\
&=25\int(\sec^3\theta-\sec\theta)d\theta \\
&=25\int\sec^3\theta d\theta-25\int\sec\theta d\theta
\end{align*}
Both integrals are standard:
\begin{align*}
&=25\left(\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|\right)-25\ln|\sec\theta+\tan\theta|+C \\
&=\frac{25}{2}\sec\theta\tan\theta+\frac{25}{2}\ln|\sec\theta+\tan\theta|-25\ln|\sec\theta+\tan\theta|+C \\
&=\frac{25}{2}\sec\theta\tan\theta-\frac{25}{2}\ln|\sec\theta+\tan\theta|+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\frac{25}{2}\cdot\frac{\sqrt{x^2+25}}{5}\cdot\frac{x}{5}-\frac{25}{2}\ln\left|\frac{\sqrt{x^2+25}}{5}+\frac{x}{5}\right|+C \\
&=\frac{1}{2}x\sqrt{x^2+25}-\frac{25}{2}\ln\left|\sqrt{x^2+25}+x\right|+C
\end{align*}
\subsection*{Problem 12}
$$\int\cos\left(x\right)\sqrt{4-\sin^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\sqrt{4-u^2}du$$
This integral is standard:
$$=2\arcsin\left(\frac{u}{2}\right)+\frac{1}{2}u\sqrt{4-u^2}+C$$
Undoing the substitution(s):
$$=2\arcsin\left(\frac{\sin(x)}{2}\right)+\frac{1}{2}\sin(x)\sqrt{4-\sin^2(x)}+C$$
\subsection*{Problem 13}
$$\int\frac{1}{x^2-x+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{x^2-x+\frac{1}{4}+\frac{3}{4}}dx \\
&=4\int\frac{1}{4x^2-4x+1+3}dx \\
&=4\int\frac{1}{(2x-1)^2+3}
\end{align*}
Substituting
\begin{align*}
u&=2x-1 \\
du&=2dx
\end{align*}
yields:
$$=2\int\frac{1}{u^2+3}du$$
This integral is standard:
$$=\frac{2}{\sqrt{3}}\arctan\left(\frac{u}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
$$=\frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 14}
$$\int\sqrt{x^2+x+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{x^2+x+\frac{1}{4}+\frac{3}{4}}dx \\
&=\frac{1}{2}\int\sqrt{4x^2+4x+1+3}dx \\
&=\frac{1}{2}\int\sqrt{(2x+1)^2+3}dx
\end{align*}
Substituting:
\begin{align*}
u&=2x+1 \\
du&=2dx
\end{align*}
yields:
$$=\frac{1}{4}\int\sqrt{u^2+3}du$$
This integral is standard:
\begin{align*}
&=\frac{1}{4}\left(\frac{1}{2}u\sqrt{u^2+3}+\frac{3}{2}\ln\left|\sqrt{u^2+3}+u\right|\right)+C \\
&=\frac{1}{8}u\sqrt{u^2+3}+\frac{3}{8}\ln\left|\sqrt{u^2+3}+u\right|+C
\end{align*}
Undoing the substitution(s):
$$=\frac{2x+1}{8}\sqrt{4x^2+4x+4}+\frac{3}{8}\ln\left|\sqrt{4x^2+4x+4}+2x+1\right|+C$$
\subsection*{Problem 15}
$$\int\frac{5x+31}{3x^2-4x+11}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{5x-\frac{10}{3}+\frac{103}{3}}{3x^2-4x+11}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+\frac{103}{9}\int\frac{1}{x^2-\frac{4}{3}x+\frac{11}{3}}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+\frac{103}{9}\int\frac{1}{x^2-\frac{4}{3}x+\frac{4}{9}+\frac{29}{9}}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+103\int\frac{1}{9x^2-12x+4+29}dx \\
&=\frac{5}{6}\int\frac{6x-4}{3x^2-4x+11}dx+103\int\frac{1}{(3x-2)^2+29}
\end{align*}
Substituting:
\begin{align*}
u&=3x^2-4x+11 &v&=3x-2 \\
du&=(6x-4)dx &dv&=3dx
\end{align*}
yields:
$$=\frac{5}{6}\int\frac{1}{u}+\frac{103}{3}\int\frac{1}{v^2+29}dv$$
Both integrals are standard:
$$=\frac{5}{6}\ln|u|+\frac{103}{3\sqrt{29}}\arctan\left(\frac{v}{\sqrt{29}}\right)+C$$
Undoing the substitution(s):
$$=\frac{5}{6}\ln\left|3x^2-4x+11\right|+\frac{103}{3\sqrt{29}}\arctan\left(\frac{3x-2}{\sqrt{29}}\right)+C$$
\subsection*{Problem 16}
$$\int\frac{x^4+1}{x^2+2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4+2x^2-2x^2-4+5}{x^2+2}dx \\
&=\int\frac{x^2(x^2+2)-2(x^2+2)+5}{x^2+2}dx \\
&=\int\left(x^2-2+\frac{5}{x^2+2}\right)dx \\
&=\int x^2dx-2\int dx+5\int\frac{1}{x^2+2}dx
\end{align*}
All three integrals are standard:
$$=\frac{1}{3}x^3-2x+\frac{5}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)+C$$
\subsection*{Problem 17}
$$\int\frac{1}{5+4\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
du&=\frac{1+u^2}{2}dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=2\int\frac{1}{5+4\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{1}{1+u^2}du \\
&=2\int\frac{1}{5(1+u^2)+4(1-u^2)}du \\
&=2\int\frac{1}{5+5u^2+4-4u^2}du \\
&=2\int\frac{1}{9+u^2}du
\end{align*}
This integral is standard:
$$=\frac{2}{3}\arctan\left(\frac{u}{3}\right)+C$$
Undoing the substitution(s):
$$=\frac{2}{3}\arctan\left(\frac{\tan(\frac{x}{2})}{3}\right)+C$$
\subsection*{Problem 18}
$$\int\frac{\sqrt{x}}{1+x}dx$$
Substituting
\begin{align*}
\tan\theta&=\sqrt{x} \\
\tan^2\theta&=x \\
2\tan\theta\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=2\int\frac{\tan\theta}{1+\tan^2\theta}\tan\theta\sec^2\theta d\theta \\
&=2\int\frac{\tan^2\theta\sec^2\theta}{\sec^2\theta}d\theta \\
&=2\int\left(\sec^2\theta-1\right)d\theta \\
&=2\int\sec^2\theta-2\int d\theta
\end{align*}
Both integrals are standard:
$$=2\tan\theta-2\theta+C$$
Undoing the substitution(s):
$$=2\sqrt{x}-2\arctan\left(\sqrt{x}\right)+C$$
\subsection*{Problem 19}
$$\int\frac{\cos\left(x\right)}{4-\sin^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\frac{1}{4-u^2}du$$
This integral is standard:
$$=\frac{1}{4}\ln\left|\frac{2+u}{2-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{4}\ln\left|\frac{2+\sin(x)}{2-\sin(x)}\right|+C$$
\subsection*{Problem 20}
$$\int\frac{\cos\left(2x\right)}{\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{2\cos^2(x)-1}{\cos(x)}dx \\
&=\int(2\cos(x)-\sec(x))dx \\
&=2\int\cos(x)dx-\int\sec(x)dx
\end{align*}
Both integrals are standard:
$$=2\sin(x)-\ln|\sec(x)+\tan(x)|+C$$
\subsection*{Problem 21}
$$\int\frac{\tan\left(x\right)}{\ln\left(\cos\left(x\right)\right)}dx$$
Substituting
\begin{align*}
u&=-\ln(\cos(x)) \\
du&=\tan(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln|u|+C$$
Undoing the substitution(s):
$$=-\ln\left|-\ln(\cos(x))\right|+C$$
\subsection*{Problem 22}
$$\int\frac{x^7}{\sqrt{1-x^4}}dx$$
Rewriting the integral:
$$=\int\frac{x^3\cdot x^4}{\sqrt{1-x^4}}dx$$
Substituting:
\begin{align*}
u&=1-x^4 \\
1-u&=x^4 \\
-du&=4x^3dx
\end{align*}
Yields:
\begin{align*}
&=-\frac{1}{4}\int\frac{1-u}{\sqrt{u}}du \\
&=-\frac{1}{4}\int\left(u^{-1/2}-\sqrt{u}\right)du \\
&=-\frac{1}{4}\int u^{-1/2}du+\frac{1}{4}\int\sqrt{u}du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{2}\sqrt{u}+\frac{1}{6}u^{3/2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\sqrt{1-x^4}+\frac{1}{6}(1-x^4)^{3/2}+C$$
\subsection*{Problem 23}
$$\int\ln\left(1+x\right)dx$$
Substituting
\begin{align*}
u&=1+x \\
du&=dx
\end{align*}
Yields:
$$=\int\ln(u)du$$
This integral is standard:
$$=u\ln(u)-u+C$$
Undoing the substitution(s):
$$=(1+x)\ln(1+x)-(1+x)+C$$
\subsection*{Problem 24}
$$\int x\arcsec\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\arcsec(x) &v&=\frac{1}{2}x^2 \\
du&=\frac{1}{x\sqrt{x^2-1}}dx &dv&=xdx
\end{align*}
yields:
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\int\frac{x}{\sqrt{x^2-1}}dx$$
Substituting
\begin{align*}
w&=x^2-1 \\
dw&=2xdx
\end{align*}
yields:
$$=-\frac{1}{4}\int\frac{1}{\sqrt{w}}dw$$
This integral is standard:
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\sqrt{w}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}x^2\arcsec(x)-\frac{1}{2}\sqrt{x^2-1}+C$$
\subsection*{Problem 25}
$$\int\sqrt{x^2+9}dx$$
This integral is standard:
$$=\frac{1}{2}x\sqrt{x^2+9}+\frac{9}{2}\ln\left|\sqrt{x^2+9}+x\right|+C$$
\subsection*{Problem 26}
$$\int\frac{x^2}{\sqrt{4-x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2}{\sqrt{4(1-\frac{x^2}{4})}}dx \\
&=\frac{1}{2}\int\frac{x^2}{\sqrt{1-(\frac{x}{2})^2}}dx
\end{align*}
Substituting
\begin{align*}
\sin\theta&=\frac{x}{2} \\
2\sin\theta&=x \\
2\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(2\sin\theta)^2\cos\theta}{\sqrt{1-\sin^2\theta}}d\theta \\
&=4\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}d\theta \\
&=4\int\sin^2\theta d\theta \\
&=4\int\left(\frac{1-\cos2\theta}{2}\right)d\theta \\
&=2\int(1-\cos2\theta)d\theta \\
&=2\int d\theta-2\int\cos2\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=2\theta \\
du&=2d\theta
\end{align*}
yields:
$$=2\int d\theta-\int\cos(u)du$$
Both integrals are standard:
$$=2\theta-\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=2\theta-\sin(2\theta)+C \\
&=2\theta-2\sin\theta\cos\theta+C \\
&=2\arcsin\left(\frac{x}{2}\right)-2\cdot\frac{x}{2}\cdot\frac{\sqrt{4-x^2}}{2}+C \\
&=2\arcsin\left(\frac{x}{2}\right)-\frac{1}{2}x\sqrt{4-x^2}+C
\end{align*}
\subsection*{Problem 27}
$$\int\sqrt{2x-x^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1-1+2x-x^2}dx \\
&=\int\sqrt{1-(1-2x+x^2)}dx \\
&=\int\sqrt{1-(1-x)^2}dx
\end{align*}
Substituting:
\begin{align*}
u&=1-x \\
du&=-dx
\end{align*}
yields:
$$=-\int\sqrt{1-u^2}du$$
This integral is standard:
$$=-\frac{1}{2}\arcsin(u)-\frac{1}{2}u\sqrt{1-u^2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\arcsin(1-x)-\frac{1}{2}(1-x)\sqrt{2x-x^2}+C$$
\subsection*{Problem 28}
$$\int\frac{4x-2}{x^3-x}dx$$
Factoring the integral:
$$=\int\frac{4x-2}{x(x-1)(x+1)}dx$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{2}{x}+\frac{1}{x-1}+\frac{-3}{x+1}\right)dx \\
&=2\int\frac{1}{x}dx+\int\frac{1}{x-1}dx-3\int\frac{1}{x+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x-1 &v&=x+1 \\
du&=dx &dv&=dx
\end{align*}
yields:
$$2\int\frac{1}{x}dx+\int\frac{1}{u}du-3\int\frac{1}{v}dv$$
All integrals are standard:
\begin{align*}
&=2\ln|x|+\ln|u|-3\ln|v|+C \\
&=\ln\left|\frac{x^2u}{v^3}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{x^2(x-1)}{(x+1)^3}\right|+C$$
\subsection*{Problem 29}
$$\int\frac{x^4}{x^2-2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4-2x^2+2x^2-4+4}{x^2-2}dx \\
&=\int\left(\frac{x^4-2x^2}{x^2-2}+\frac{2x^2-4}{x^2-2}+\frac{4}{x^2-2}\right)dx \\
&=\int\frac{x^2(x^2-2)}{x^2-2}dx+\int\frac{2(x^2-2)}{x^2-2}dx+4\int\frac{1}{x^2-2}dx \\
&=\int x^2dx+2\int dx+4\int\frac{1}{x^2-2}dx
\end{align*}
All three integrals are standard:
$$=\frac{1}{3}x^3+2x+\sqrt{2}\ln\left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C$$
\subsection*{Problem 30}
$$\int\frac{\sec\left(x\right)\tan\left(x\right)}{\sec\left(x\right)+\sec^2\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\tan(x)}{1+\sec(x)}\cdot\frac{\cos(x)}{\cos(x)}dx \\
&=\int\frac{\sin(x)}{\cos(x)+1}dx
\end{align*}
Substituting
\begin{align*}
u&=\cos(x)+1 \\
du&=-\sin(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln\left|u\right|+C$$
Undoing the substitution(s):
$$=-\ln\left|\cos(x)+1\right|+C$$
\subsection*{Problem 31}
$$\int\frac{x}{\left(x^2+2x+2\right)^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x}{(x^2+2x+1+1)^2}dx \\
&=\int\frac{x}{((x+1)^2+1)^2}dx
\end{align*}
Substituting
\begin{align*}
\tan\theta&=x+1 \\
\tan\theta-1&=x \\
\sec^2\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\tan\theta-1}{(\tan^2\theta+1)^2}\sec^2\theta d\theta \\
&=\int\frac{\tan\theta-1}{\sec^2\theta}\cdot\frac{\cos^2\theta}{\cos^2\theta}d\theta \\
&=\int\left(\sin\theta\cos\theta-\cos^2\theta\right) \\
&=\int\left(\sin\theta\cos\theta-\frac{1+\cos2\theta}{2}\right)d\theta \\
&=\int\sin\theta\cos\theta d\theta-\frac{1}{2}\int d\theta-\frac{1}{2}\int\cos2\theta d\theta
\end{align*}
Substituting:
\begin{align*}
u&=\sin\theta &v&=2\theta \\
du&=\cos\theta d\theta &dv&=2d\theta
\end{align*}
yields:
$$=\int udu-\frac{1}{2}\int d\theta-\frac{1}{4}\int\cos(u)du$$
All three integrals are standard:
$$=\frac{1}{2}u^2-\frac{1}{2}\theta-\frac{1}{4}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{2}\sin^2\theta-\frac{1}{2}\theta-\frac{1}{4}\sin2\theta+C \\
&=\frac{1}{2}\sin^2\theta-\frac{1}{2}\theta-\frac{1}{2}\sin\theta\cos\theta+C \\
&=\frac{1}{2}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right)^2-\frac{1}{2}\arctan(x+1)-\frac{1}{2}\cdot\frac{x+1}{\sqrt{x^2+2x+2}}\cdot\frac{1}{\sqrt{x^2+2x+2}}+C \\
&=\frac{x^2+2x+1}{2(x^2+2x+2)}-\frac{x+1}{2(x^2+2x+2)}-\frac{1}{2}\arctan(x+1)+C \\
&=\frac{x^2+x}{2x^2+4x+4}-\frac{1}{2}\arctan(x+1)+C
\end{align*}
\subsection*{Problem 32}
$$\int\frac{x^{1/3}}{x^{1/2}+x^{1/4}}dx$$
Substituting
\begin{align*}
u^3&=\sqrt[4]{x} \\
u^4&=\sqrt[3]{x} \\
u^6&=\sqrt{x} \\
u^{12}&=x\\
12u^{11}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{u^4}{u^6+u^3}12u^{11}du \\
&=12\int\frac{u^{12}}{u^3+1}du
\end{align*}
Performing long division on the integral yields:
\begin{align*}
&=12\int\left(u^9-u^6+u^3-1+\frac{1}{u^3+1}\right)du \\
&=12\int u^9du-12\int u^6du+12\int u^3du-12\int du+12\int\frac{1}{u^3+1}du
\end{align*}
Factoring the last integral:
$$=12\int\frac{1}{(u+1)(u^2-u+1)}du$$
Using partial fraction decomposition
$$=12\int\left(\frac{A}{u+1}+\frac{Bu+C}{u^2-u+1}\right)du$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=12\int\left(\frac{\frac{1}{3}}{u+1}+\frac{-\frac{1}{3}u+\frac{2}{3}}{u^2-u+1}\right)du \\
&=4\int\frac{1}{u+1}du-4\int\frac{u-2}{u^2-u+1}du
\end{align*}
Rewriting the second integral:
\begin{align*}
&=4\int\frac{1}{u+1}du-4\int\frac{u-\frac{1}{2}-\frac{3}{2}}{u^2-u+1}du \\
&=4\int\frac{1}{u+1}du-4\int\frac{u-\frac{1}{2}}{u^2-u+1}du+4\int\frac{\frac{3}{2}}{u^2-u+1}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+6\int\frac{1}{u^2-u+\frac{1}{4}+\frac{3}{4}}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+24\int\frac{1}{4u^2-4u+1+3}du \\
&=4\int\frac{1}{u+1}du-2\int\frac{2u-1}{u^2-u+1}du+24\int\frac{1}{(2u-1)^2+3}du
\end{align*}
Substituting
\begin{align*}
v&=u+1 &w&=u^2-u+1 &y&=2u-1 \\
dv&=du &dw&=(2u-1)du &dy&=2du
\end{align*}
yields:
$$=4\int\frac{1}{v}dv-2\int\frac{1}{w}dw+12\int\frac{1}{y^2+3}dy$$
All three integrals are standard:
\begin{align*}
&=4\ln|v|-2\ln|w|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C \\
&=\ln\left|v^4\right|-\ln\left|w^2\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C \\
&=\ln\left|\frac{v^4}{w^2}\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C
\end{align*}
The previous four integrals are also standard:
$$=\frac{6}{5}u^{10}-\frac{12}{7}u^7+3u^4-12u+\ln\left|\frac{v^4}{w^2}\right|+4\sqrt{3}\arctan\left(\frac{y}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{6}{5}u^{10}-\frac{12}{7}u^7+3u^4-12u+\ln\left|\frac{(u+1)^4}{(u^2-u+1)^2}\right|+4\sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right)+C \\
&=\frac{6}{5}x^{5/6}-\frac{12}{7}x^{7/12}+3x^{1/3}-12x^{1/12}+\cdots \\
&+\ln\left|\frac{(x^{1/12}+1)^4}{(x^{1/6}-x^{1/12}+1)^2}\right|+4\sqrt{3}\arctan\left(\frac{2x^{1/12}-1}{\sqrt{3}}\right)+C
\end{align*}
\subsection*{Problem 33}
$$\int\frac{1}{1+\cos\left(2x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{1+2\cos^2(x)-1}dx \\
&=\frac{1}{2}\int\sec^2(x)dx
\end{align*}
This integral is standard:
$$=\frac{1}{2}\tan(x)+C$$
\subsection*{Problem 34}
$$\int\frac{\sec\left(x\right)}{\tan\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sec(x)}{\tan(x)}\cdot\frac{\cos(x)}{\cos(x)}dx \\
&=\int\frac{1}{\sin(x)}dx \\
&=\int\csc(x)dx
\end{align*}
This integral is standard:
$$=-\ln\left|\cot(x)+\csc(x)\right|+C$$
\subsection*{Problem 35}
$$\int\sec^3\left(x\right)\tan^3\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\sec^2(x)\tan^2(x)\sec(x)\tan(x)dx \\
&=\int\sec^2(x)(\sec^2(x)-1)\sec(x)\tan(x) \\
&=\int(\sec^4(x)-\sec^2(x))\sec(x)\tan(x)
\end{align*}
Substituting
\begin{align*}
u&=\sec(x) \\
du&=\sec(x)\tan(x)dx
\end{align*}
yields:
\begin{align*}
&=\int\left(u^4-u^2\right)du \\
&=\int u^4du-\int u^2du
\end{align*}
Both integrals are standard:
$$=\frac{1}{5}u^5-\frac{1}{3}u^3+C$$
Undoing the substitution(s):
$$=\frac{1}{5}\sec^5(x)-\frac{1}{3}\sec^3(x)+C$$
\subsection*{Problem 36}
$$\int x^2\arctan\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=\frac{1}{3}x^3 \\
du&=\frac{1}{x^2+1}dx &dv&=x^2dx
\end{align*}
yields:
$$=\frac{1}{3}x^3\arctan(x)-\frac{1}{3}\int\frac{x^3}{x^2+1}dx$$
\subsection*{Problem 37}
$$\int x\ln^3\left(x\right)dx$$
Substituting
\begin{align*}
u&=\ln(x) \\
e^u&=x \\
e^udu&=dx
\end{align*}
yields:
$$=\int u^3e^{2u}du$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $u^3$ & $e^{2u}$ \\
$-$ & $3u^2$ & $\frac{1}{2}e^{2u}$ \\
$+$ & $6u$ & $\frac{1}{4}e^{2u}$ \\
$-$ & $6$ & $\frac{1}{8}e^{2u}$ \\
$+$ & $0$ & $\frac{1}{16}e^{2u}$ \\
\hline
\end{tabular}
\end{center}
yields:
$$=\frac{1}{2}u^3e^{2u}-\frac{3}{4}u^2e^{2u}+\frac{3}{4}ue^{2u}-\frac{3}{8}e^{2u}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}x^2\ln^3(x)-\frac{3}{4}x^2\ln^2(x)+\frac{3}{4}x^2\ln(x)-\frac{3}{8}x^2+C$$
\subsection*{Problem 38}
$$\int\frac{1}{x\sqrt{1+x^2}}dx$$
Substituting
\begin{align*}
\tan\theta&=x \\
\sec^2\theta d\theta&dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\sec^2\theta}{\tan\theta\sqrt{1+\tan^2\theta}}d\theta \\
&=\int\frac{\sec^2\theta}{\tan\theta\sqrt{\sec^2\theta}}d\theta \\
&=\int\frac{\sec\theta}{\tan\theta}d\theta \\
&=\int\frac{\sec\theta}{\tan\theta}\cdot\frac{\cos\theta}{\cos\theta}d\theta \\
&=\int\frac{1}{\sin\theta}d\theta \\
&=\int\csc\theta d\theta
\end{align*}
This integral is standard:
$$=-\ln\left|\cot\theta+\csc\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\ln\left|\frac{1}{x}+\frac{\sqrt{x^2+1}}{x}\right|+C \\
&=-\ln\left|\frac{1+\sqrt{x^2+1}}{x}\right|+C \\
&=\ln\left|\frac{x}{1+\sqrt{x^2+1}}\right|+C
\end{align*}
\subsection*{Problem 39}
$$\int e^x\sqrt{1+e^{2x}}dx$$
Substituting
\begin{align*}
\tan\theta&=e^x \\
\sec^2\theta d\theta&=e^xdx
\end{align*}
yields:
\begin{align*}
&=\int\sec^2\theta\sqrt{1+\tan^2\theta}d\theta \\
&=\int\sec^2\theta\sqrt{\sec^2\theta}d\theta \\
&=\int\sec^3\theta d\theta
\end{align*}
This integral is standard:
$$=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{2}e^x\sqrt{1+e^{2x}}+\frac{1}{2}\ln\left|\sqrt{1+e^{2x}}+e^x\right|+C
\end{align*}
\subsection*{Problem 40}
$$\int\frac{x}{4x-x^2}dx$$
Rewriting the integral:
$$=\int\frac{1}{4-x}dx$$
Substituting
\begin{align*}
u&=4-x \\
du&=-dx
\end{align*}
yields:
$$=-\int\frac{1}{u}du$$
This integral is standard:
$$=-\ln\left|u\right|+C$$
Undoing the substitution(s):
$$=-\ln\left|4-x\right|+C$$
\subsection*{Problem 41}
$$\int\frac{1}{x^3\sqrt{x^2-9}}dx$$
Rewriting the integral:
\begin{align*}
=&\int\frac{1}{x^3\sqrt{9\left(\frac{x^2}{9}-1\right)}}dx \\
=&\frac{1}{3}\int\frac{1}{x^3\sqrt{\left(\frac{x}{3}\right)^2-1}}dx
\end{align*}
Substituting
\begin{align*}
\sec\theta&=\frac{x}{3} \\
3\sec\theta&=x \\
3\sec\theta\tan\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{3}\int\frac{3\sec\theta\tan\theta}{(3\sec\theta)^3\sqrt{\sec^2\theta-1}}d\theta \\
&=\frac{1}{27}\int\frac{\tan\theta}{\sec^2\theta\sqrt{\tan^2\theta}}d\theta \\
&=\frac{1}{27}\int\cos^2\theta d\theta \\
&=\frac{1}{54}\int\left(1+\cos2\theta\right)d\theta \\
&=\frac{1}{54}\int d\theta+\frac{1}{54}\int\cos2\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=2\theta \\
du&2d\theta
\end{align*}
yields:
$$=\frac{1}{54}\int d\theta+\frac{1}{108}\int\cos(u)du$$
Both integrals are standard:
$$=\frac{1}{54}\theta+\frac{1}{108}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{54}\theta+\frac{1}{108}\sin2\theta+C \\
&=\frac{1}{54}\theta+\frac{1}{54}\sin\theta\cos\theta+C \\
&=\frac{1}{54}\arcsec\left(\frac{x}{3}\right)+\frac{1}{54}\cdot\frac{\sqrt{x^2-9}}{x}\cdot\frac{3}{x}+C \\
&=\frac{1}{54}\arcsec\left(\frac{x}{3}\right)+\frac{\sqrt{x^2-9}}{18x^2}+C
\end{align*}
\subsection*{Problem 42}
$$\int\frac{x}{\left(7x+1\right)^{17}}dx$$
Substituting
\begin{align*}
u&=7x+1 \\
\frac{1}{7}u-\frac{1}{7}&=x \\
\frac{1}{7}du&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{7}\int\frac{\frac{1}{7}u-\frac{1}{7}}{u^{17}}du \\
&=\frac{1}{49}\int\frac{u-1}{u^{17}}du \\
&=\frac{1}{49}\int\left(u^{-16}-u^{-17}\right)du \\
&=\frac{1}{49}\int u^{-16}du-\frac{1}{49}\int u^{-17}du
\end{align*}
Both integrals are standard:
$$=-\frac{1}{735u^{15}}+\frac{1}{784u^{16}}+C$$
Undoing the substitution(s):
$$=-\frac{1}{735(7x+1)^{15}}+\frac{1}{784(7x+1)^{16}}+C$$
\subsection*{Problem 43}
$$\int\frac{4x^2+x+1}{4x^3+x}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{4x^2+1+x}{x(4x^2+1)}dx \\
&=\int\left(\frac{4x^2+1}{x(4x^2+1)}+\frac{x}{x(4x^2+1)}\right)dx \\
&=\int\left(\frac{1}{x}+\frac{1}{4x^2+1}\right)dx \\
&=\int\frac{1}{x}dx+\int\frac{1}{4x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=2x \\
du&=2dx
\end{align*}
yields:
$$=\int\frac{1}{x}dx+\frac{1}{2}\int\frac{1}{u^2+1}du$$
Both integrals are standard:
$$=\ln|x|+\frac{1}{2}\arctan(u)+C$$
Undoing the substitution(s):
$$=\ln\left|x\right|+\frac{1}{2}\arctan(2x)+C$$
\subsection*{Problem 44}
$$\int\frac{4x^3-x+1}{x^3+1}dx$$
Performing long division on the integral yields:
\begin{align*}
&=\int\left(4-\frac{x+3}{x^3+1}\right)dx \\
&=4\int dx-\int\frac{x+3}{(x+1)(x^2-x+1)}dx
\end{align*}
Using partial fraction decomposition
$$=-\int\left(\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\right)$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=-\int\left(\frac{\frac{2}{3}}{x+1}+\frac{-\frac{2}{3}x+\frac{7}{3}}{x^2-x+1}\right)dx
\\
&=-\frac{1}{3}\int\left(\frac{2}{x+1}-\frac{2x-7}{x^2-x+1}\right)dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-7}{x^2-x+1}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1-6}{x^2-x+1}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\left(\frac{2x-1}{x^2-x+1}-\frac{6}{x^2-x+1}\right)dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-2\int\frac{1}{x^2-x+\frac{1}{4}+\frac{3}{4}}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-8\int\frac{1}{4x^2-4x+1+3}dx \\
&=-\frac{2}{3}\int\frac{1}{x+1}dx+\frac{1}{3}\int\frac{2x-1}{x^2-x+1}dx-8\int\frac{1}{(2x-1)^2+3}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 &v&=x^2-x+1 &w&=2x-1 \\
du&=dx &dv&=(2x-1)dx &dw&=2dx
\end{align*}
yields:
$$=-\frac{2}{3}\int\frac{1}{u}du+\frac{1}{3}\int\frac{1}{v}dv-4\int\frac{1}{w^2+3}dw$$
All three integrals are standard:
\begin{align*}
&=-\frac{2}{3}\ln|u|+\frac{1}{3}\ln|v|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C \\
&=-\frac{1}{3}\ln\left|u^2\right|+\frac{1}{3}\ln|v|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C \\
&=\frac{1}{3}\ln\left|\frac{v}{u^2}\right|-\frac{4}{\sqrt{3}}\arctan\left(\frac{w}{\sqrt{3}}\right)+C
\end{align*}
Undoing the substitution(s):
$$=4x+\frac{1}{3}\ln\left|\frac{x^2-x+1}{(x+1)^2}\right|-\frac{4}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 45}
$$\int\tan^2\left(x\right)\sec\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int(\sec^2(x)-1)\sec(x)dx \\
&=\int(\sec^3(x)-\sec(x))dx \\
&=\int\sec^3(x)dx-\int\sec(x)dx
\end{align*}
Both integrals are standard:
\begin{align*}
&=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln|\sec(x)+\tan(x)-\ln|\sec(x)+\tan(x)|+C \\
&=\frac{1}{2}\sec(x)\tan(x)-\frac{1}{2}\ln|\sec(x)-\tan(x)|+C
\end{align*}
\subsection*{Problem 46}
$$\int\frac{x^2+2x+2}{\left(x+1\right)^3}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2+2x+1+1}{(x+1)^3}dx \\
&=\int\left(\frac{x^2+2x+1}{(x+1)^3}+\frac{1}{(x+1)^3}\right)dx \\
&=\int\left(\frac{(x+1)^2}{(x+1)^3}+\frac{1}{(x+1)^3}\right)dx \\
&=\int\frac{1}{x+1}dx+\int\frac{1}{(x+1)^3}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 \\
du&=dx
\end{align*}
Yields:
$$=\int\frac{1}{u}du+\int\frac{1}{u^3}du$$
Both integrals are standard:
$$=\ln|u|-\frac{1}{2u^2}+C$$
Undoing the substitution(s):
$$=\ln|x+1|-\frac{1}{2(x+1)^2}+C$$
\subsection*{Problem 47}
$$\int\frac{x^4+2x+2}{x^5+x^4}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^4+2(x+1)}{x^5+x^4}dx \\
&=\int\left(\frac{x^4}{x^4(x+1)}+\frac{2(x+1)}{x^4(x+1)}\right)dx \\
&=\int\frac{1}{x+1}dx+2\int\frac{1}{x^4}dx
\end{align*}
Substituting
\begin{align*}
u&=x+1 \\
du&=dx
\end{align*}
yields:
$$=\int\frac{1}{u}du+2\int\frac{1}{x^4}dx$$
Both integrals are standard:
$$=\ln|u|-\frac{2}{3x^3}+C$$
Undoing the substitution(s):
$$=\ln|x+1|-\frac{2}{3x^3}+C$$
\subsection*{Problem 48}
$$\int\frac{8x^2-4x+7}{\left(x^2+1\right)\left(4x+1\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{8x^2+8-4x-1}{(x^2+1)(4x+1)}dx \\
&=\int\frac{8(x^2+1)-(4x+1)}{(x^2+1)(4x+1)} \\
&=\int\left(\frac{8(x^2+1)}{(x^2+1)(4x+1)}-\frac{4x+1}{(x^2+1)(4x+1)}\right)dx \\
&=\int\left(\frac{8}{4x+1}-\frac{1}{x^2+1}\right)dx \\
&=8\int\frac{1}{4x+1}-\int\frac{1}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=4x+1 \\
du&=4dx
\end{align*}
yields:
$$=2\int\frac{1}{u}du-\int\frac{1}{x^2+1}dx$$
Both integrals are standard:
$$=2\ln|u|-\arctan(x)+C$$
Undoing the substitution(s):
$$=2\ln|4x+1|-\arctan(x)+C$$
\subsection*{Problem 49}
$$\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{\left(x^3-1\right)^2}dx$$
Factoring the integral:
\begin{align*}
&=\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{((x-1)(x^2+x+1))^2}dx \\
&=\int\frac{3x^5-x^4+2x^3-12x^2-2x+1}{(x-1)^2(x^2+x+1)^2}dx
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+x+1}+\frac{Ex+F}{(x^2+x+1)^2}\right)dx$$
where $A,B,C,D,E,F\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{x-1}-\frac{1}{(x-1)^2}+\frac{2x+1}{x^2+x+1}+\frac{4x+2}{(x^2+x+1)^2}\right)dx \\
&=\int\frac{1}{x-1}dx-\int\frac{1}{(x-1)^2}dx+\int\frac{2x+1}{x^2+x+1}dx+2\int\frac{2x+1}{(x^2+x+1)^2}dx
\end{align*}
Substituting
\begin{align*}
u&=x-1 &v&=x^2+x+1 \\
du&=dx &dv&=(2x+1)dx
\end{align*}
yields:
$$=\int\frac{1}{u}du-\int\frac{1}{u^2}du+\int\frac{1}{v}dv+2\int\frac{1}{v^2}dv$$
All four integrals are standard:
\begin{align*}
&=\ln|u|+\frac{1}{u}+\ln|v|-\frac{2}{v}+C \\
&=\ln|uv|+\frac{v}{uv}-\frac{2u}{uv}+C \\
&=\ln|uv|+\frac{v-2u}{uv}+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\ln|(x-1)(x^2+x+1)|+\frac{x^2+x+1-2(x-1)}{(x-1)(x^2+x+1)}+C \\
&=\ln\left|x^3-1\right|+\frac{x^2-x+3}{x^3-1}+C
\end{align*}
\subsection*{Problem 50}
$$\int\frac{x}{x^4+4x^2+8}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x}{x^4+4x^2+4+4}dx \\
&=\int\frac{x}{(x^2+2)^2+4}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+2 \\
du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u^2+4}du$$
This integral is standard:
$$=\frac{1}{4}\arctan\left(\frac{u}{2}\right)+C$$
Undoing the substitution(s):
$$=\frac{1}{4}\arctan\left(\frac{x^2+2}{2}\right)+C$$
\subsection*{Problem 51}
$$\int\frac{1}{4+5\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{4+5\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{4(1+u^2)+5(1-u^2)}du \\
&=2\int\frac{1}{4+4u^2+5-5u^2}du \\
&=2\int\frac{1}{9-u^2}du
\end{align*}
This integral is standard:
$$=\frac{1}{3}\ln\left|\frac{3+u}{3-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\ln\left|\frac{3+\tan(\frac{x}{2})}{3-\tan(\frac{x}{2})}\right|+C$$
\subsection*{Problem 52}
$$\int\frac{\left(1+x^{2/3}\right)^{3/2}}{x^{1/3}}dx$$
Substituting
\begin{align*}
u&=1+x^{2/3} \\
du&=\frac{2}{3}x^{-1/3}dx
\end{align*}
yields:
$$=\frac{3}{2}\int u^{3/2}du$$
This integral is standard:
$$=\frac{3}{5}u^{5/2}+C$$
Undoing the substitution(s):
$$=\frac{3}{5}(1+x^{2/3})^{5/2}+C$$
\subsection*{Problem 53}
$$\int\frac{\arcsin^2\left(x\right)}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=\arcsin(x) \\
du&=\frac{1}{\sqrt{1-x^2}}dx
\end{align*}
yields:
$$=\int u^2du$$ \\
This integral is standard:
$$=\frac{1}{3}u^3+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\arcsin^3(x)+C$$
\subsection*{Problem 54}
$$\int\frac{1}{x^{3/2}\left(1+x^{1/3}\right)}dx$$
Substituting
\begin{align*}
u&=x^{1/6} \\
u^2&=x^{1/3} \\
u^6&=x \\
u^9&=x^{3/2} \\
6u^5du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{u^9(1+u^2)}6u^5du \\
&=6\int\frac{1}{u^4(1+u^2)}du
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u^3}+\frac{D}{u^4}+\frac{Eu+F}{u^2+1}\right)du$$
where $A,B,C,D,E,F\in\mathbb{R}$ yields:
\begin{align*}
&=6\int\left(-\frac{1}{u^2}+\frac{1}{u^4}+\frac{1}{u^2+1}\right)du \\
&=-6\int\frac{1}{u^2}du+6\int\frac{1}{u^4}du+6\int\frac{1}{u^2+!}du
\end{align*}
All three integrals are standard:
$$=\frac{6}{u}-\frac{2}{u^3}+6\arctan(u)+C$$
Undoing the substitution(s):
$$=\frac{6}{x^{1/6}}-\frac{2}{\sqrt{x}}+6\arctan\left(x^{1/6}\right)+C$$
\subsection*{Problem 55}
$$\int\tan^3\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int\tan^2(x)\tan(x)dx \\
&=\int(\sec^2(x)-1)\tan(x)dx \\
&=\int(\sec^2(x)\tan(x)-\tan(x))dx \\
&=\int\sec^2(x)\tan(x)dx-\int\tan(x)dx
\end{align*}
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int udu-\int\tan(x)dx$$
Both integrals are standard:
$$=\frac{1}{2}u^2+\ln|\cos(x)|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$
\subsection*{Problem 56}
$$\int\sin^2\left(x\right)\cos^4\left(x\right)dx$$
Rewriting the integral:
\begin{align*}
&=\int(1-\cos^2(x))\cos^4(x)dx \\
&=\int(\cos^4(x)-\cos^6(x))dx \\
&=\int\cos^4(x)dx-\int\cos^6(x)dx
\end{align*}
Using the reduction rule for $\cos^n(x)$, it's known that:
\begin{align*}
\int\cos^4(x)dx&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{4}\int\cos^2(x)dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{4}\int\frac{1+\cos(2x)}{2}dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\int(1+\cos(2x))dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\int dx+\frac{3}{8}\int\cos(2x)dx \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}x+\frac{3}{16}\sin(2x)+C \\
&=\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\sin(x)\cos(x)+\frac{3}{8}x+C
\end{align*}
and
$$\int\cos^6(x)dx=\frac{1}{6}\cos^5(x)\sin(x)+\frac{5}{6}\int\cos^4(x)dx$$
Rewriting the integral:
\begin{align*}
&=\int\cos^4(x)dx-\frac{1}{6}\cos^5(x)\sin(x)-\frac{5}{6}\int\cos^4(x)dx \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{6}\int\cos^4(x)dx \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{6}\left(\frac{1}{4}\cos^3(x)\sin(x)+\frac{3}{8}\sin(x)\cos(x)+\frac{3}{8}x+C\right) \\
&=-\frac{1}{6}\cos^5(x)\sin(x)+\frac{1}{24}\cos^3(x)\sin(x)+\frac{1}{16}\sin(x)\cos(x)+\frac{1}{16}x+C
\end{align*}
\subsection*{Problem 57}
$$\int\frac{xe^{x^2}}{1+e^{2x^2}}dx$$
Substituting
\begin{align*}
u&=e^{x^2} \\
du&=2xe^{x^2}
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=\frac{1}{2}\arctan(u)+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\arctan\left(e^{x^{2}}\right)+C$$
\subsection*{Problem 58}
$$\int\frac{\cos^3\left(x\right)}{\sqrt{\sin\left(x\right)}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\cos(x)\cos^2(x)}{\sqrt{\sin(x)}}dx \\
&=\int\frac{\cos(x)(1-\sin^2(x))}{\sqrt{\sin(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1-u^2}{\sqrt{u}}du \\
&=\int\left(\frac{1}{\sqrt{u}}-u^{3/2}\right)du \\
&=\int\frac{1}{\sqrt{u}}du-\int u^{3/2}du
\end{align*}
Both integrals are standard:
$$=2\sqrt{u}-\frac{2}{5}u^{5/2}+C$$
Undoing the substitution(s):
$$=2\sqrt{\sin(x)}-\frac{2}{5}\left(\sin(x)\right)^{5/2}+C$$
\subsection*{Problem 59}
$$\int x^3e^{-x^2}dx$$
Rewriting the integral:
$$=\int x\cdot x^2e^{-x^2}dx$$
Substituting
\begin{align*}
u&=-x^2 \\
-u&=x^2 \\
-du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int ue^udu$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $u$ & $e^u$ \\
$-$ & 1 & $e^u$ \\
$+$ & 0 & $e^u$ \\
\hline
\end{tabular}
\end{center}
yields:
$$=\frac{1}{2}ue^u-\frac{1}{2}e^u+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}x^2e^{-x^2}-\frac{1}{2}e^{-x^2}+C$$
\subsection*{Problem 60}
$$\int\sin\left(\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
u^2&=x \\
2udu&=dx
\end{align*}
yields:
$$=2\int u\sin(u)du$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $u$ & $\sin(u)$ \\
$-$ & 1 & $-\cos(u)$ \\
$+$ & 0 & $-\sin(u)$ \\
\hline
\end{tabular}
\end{center}
yields:
$$=-2u\cos(u)+2\sin(u)+C$$
Undoing the substitution(s):
$$=-2\sqrt{x}\cos\left(\sqrt{x}\right)+2\sin\left(\sqrt{x}\right)+C$$
\subsection*{Problem 61}
$$\int\frac{\arcsin\left(x\right)}{x^2}dx$$
Substituting
\begin{align*}
\sin\theta&=x \\
\theta&=\sin(x) \\
\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{\theta\cos\theta}{\sin^2\theta}d\theta \\
&=\int u\cot\theta\csc\theta d\theta
\end{align*}
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $\theta$ & $\cot\theta\csc\theta$ \\
$-$ & 1 & $-\csc\theta$ \\
$+$ & 0 & $\ln|\csc\theta+\cot\theta|$ \\
\hline
\end{tabular}
\end{center}
yields:
$$=-\theta\csc\theta-\ln\left|\csc\theta+\cot\theta\right|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{\arcsin(x)}{x}-\ln\left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|+C \\
&=-\frac{\arcsin(x)}{x}-\ln\left|\frac{1+\sqrt{1-x^2}}{x}\right|+C
\end{align*}
\subsection*{Problem 62}
$$\int\sqrt{x^2-9}dx$$
This integral is standard:
$$=\frac{1}{2}x\sqrt{x^2-9}-\frac{9}{2}\ln\left|\sqrt{x^2-9}+x\right|+C$$
\subsection*{Problem 63}
$$\int x^2\sqrt{1-x^2}dx$$
Substituting
\begin{align*}
\sin\theta&=x \\
\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\sin^2\theta\cos\theta\sqrt{1-\sin^2\theta}d\theta \\
&=\int\sin^2\theta\cos\theta\sqrt{\cos^2\theta}d\theta \\
&=\int\sin^2\theta\cos^2\theta d\theta \\
&=\frac{1}{4}\int4\sin^2\theta\cos^2\theta d\theta \\
&=\frac{1}{4}\int(2\sin\theta\cos\theta)^2 d\theta \\
&=\frac{1}{4}\int\sin^22\theta d\theta \\
&=\frac{1}{4}\int\left(\frac{1-\cos4\theta}{2}\right)d\theta \\
&=\frac{1}{8}\int(1-\cos4\theta)d\theta \\
&=\frac{1}{8}\int d\theta-\frac{1}{8}\int\cos4\theta d\theta
\end{align*}
Substituting
\begin{align*}
u&=4\theta \\
du&=4d\theta
\end{align*}
yields:
$$=\frac{1}{8}\int d\theta-\frac{1}{32}\int\cos(u)du$$
Both integrals are standard:
$$=\frac{1}{8}\theta-\frac{1}{32}\sin(u)+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{8}\theta-\frac{1}{32}\sin4\theta+C \\
&=\frac{1}{8}\theta-\frac{1}{16}\sin2\theta\cos2\theta+C \\
&=\frac{1}{8}\theta-\frac{1}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}x\sqrt{1-x^2}\left(\left(\sqrt{1-x^2}\right)^2-x^2\right)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}x\sqrt{1-x^2}(1-x^2-x^2)+C \\
&=\frac{1}{8}\arcsin(x)-\frac{1}{8}(x-2x^3)\sqrt{1-x^2}+C
\end{align*}
\subsection*{Problem 64}
$$\int x\sqrt{2x-x^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\left(1-\frac{1}{2}(2-2x)\right)\sqrt{2x-x^2}dx \\
&=\int\sqrt{2x-x^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-1+2x-x^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-(1-2x+x^2)}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx \\
&=\int\sqrt{1-(1-x)^2}dx-\frac{1}{2}\int(2-2x)\sqrt{2x-x^2}dx
\end{align*}
Substituting
\begin{align*}
u&=1-x &v&=2x-x^2 \\
du&=-dx &dv&=(2-2x)dx
\end{align*}
$$=-\int\sqrt{1-u^2}du-\frac{1}{2}\int\sqrt{v}dv$$
Both integrals are standard:
$$=-\frac{1}{2}\arcsin(u)-\frac{1}{2}u\sqrt{1-u^2}-\frac{1}{3}v^{3/2}+C$$
Undoing the substitution(s):
$$=-\frac{1}{2}\arcsin(1-x)-\frac{1-x}{2}\sqrt{2x-x^2}-\frac{1}{3}(2x-x^2)^{3/2}+C$$
\subsection*{Problem 65}
$$\int\frac{x-2}{4x^2+4x+1}dx$$
Rewriting the integral:
$$=\int\frac{x-2}{(2x+1)^2}dx$$
Substituting
\begin{align*}
u&=2x+1 \\
\frac{u-1}{2}&=x \\
\frac{1}{2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\frac{\frac{1}{2}u-\frac{5}{2}}{u^2}du \\
&=\frac{1}{4}\int\frac{u-5}{u^2}du \\
&=\frac{1}{4}\int\left(\frac{1}{u}-\frac{5}{u^2}\right)du \\
&=\frac{1}{4}\int\frac{1}{u}du-\frac{5}{4}\int\frac{1}{u^2}du
\end{align*}
Both integrals are standard:
$$=\frac{1}{4}\ln|u|+\frac{5}{4u}+C$$
Undoing the substitution(s):
\begin{align*}
&=\frac{1}{4}\ln|2x+1|+\frac{5}{4(2x+1)}+C \\
&=\frac{1}{4}\ln|2x+1|+\frac{5}{8x+4}+C
\end{align*}
\subsection*{Problem 66}
$$\int\frac{2x^2-5x-1}{x^3-2x^2-x+2}dx$$
Factoring the integral:
\begin{align*}
&=\int\frac{2x^2-5x-1}{x^2(x-2)-(x-2)}dx \\
&=\int\frac{2x^2-5x-1}{(x^2-1)(x-2)}dx \\
&=\int\frac{2x^2-5x-1}{(x-1)(x+1)(x-2)}dx
\end{align*}
Using partial fraction decomposition
$$=\int\left(\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{2}{x-1}+\frac{1}{x+1}+\frac{-1}{x-2}\right)dx \\
&=2\int\frac{1}{x-1}dx+\int\frac{1}{x+1}dx-\int\frac{1}{x-2}dx
\end{align*}
Substituting
\begin{align*}
u_1&=x-1 &u_2&=x+1 &u_3&=x-2 \\
du_1&=dx &du_2&=dx &du_3&=dx
\end{align*}
yields:
$$=2\int\frac{1}{u_1}du_1+\int\frac{1}{u_2}du_2-\int\frac{1}{u_3}du_3$$
All three integrals are standard:
\begin{align*}
&=2\ln\left|u_1\right|+\ln\left|u_2\right|-\ln\left|u_3\right|+C \\
&=\ln\left|u_1^2\right|+\ln\left|u_2\right|-\ln\left|u_3\right|+C \\
&=\ln\left|\frac{u_1^2u_2}{u_3}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{(x-1)^2(x+1)}{x-2}\right|+C$$
\subsection*{Problem 67}
$$\int\frac{e^{2x}}{e^{2x}-1}dx$$
Substituting
\begin{align*}
u&=e^{2x}-1 \\
du&=2e^{2x}dx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=\frac{1}{2}\ln|u|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|e^{2x}-1\right|+C$$
\subsection*{Problem 68}
$$\int\frac{\cos\left(x\right)}{\sin^2\left(x\right)-3\sin\left(x\right)+2}dx$$
Substituting
\begin{align*}
u&=\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u^2-3u+2}du$$
Factoring the integral:
$$=\int\frac{1}{(u-2)(u-1)}du$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{u-2}+\frac{B}{u-1}\right)du$$
where $A,B\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{u-2}+\frac{-1}{u-1}\right)du \\
&=\int\frac{1}{u-2}du-\int\frac{1}{u-1}du
\end{align*}
Substituting
\begin{align*}
v_1&=u-2 &v_2&=u-1 \\
dv_1&=du &dv_2&=du
\end{align*}
yields:
$$=\int\frac{1}{v_1}dv_1-\int\frac{1}{v_2}dv_2$$
Both integrals are standard:
\begin{align*}
&=\ln\left|v_1\right|-\ln\left|v_2\right|+C \\
&=\ln\left|\frac{v_1}{v_2}\right|+C
\end{align*}
Undoing the substitution(s):
$$=\ln\left|\frac{\sin(x)-2}{\sin(x)-1}\right|+C$$
\subsection*{Problem 69}
$$\int\frac{2x^3+3x^2+4}{\left(x+1\right)^4}dx$$
Substituting
\begin{align*}
u&=x+1 \\
u-1&=x \\
du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{2(u-1)^3+3(u-1)^2+4}{u^4}du \\
&=\int\frac{2(u^3-3u^2+3u-1)+3(u^2-2u+1)+4}{u^4}du \\
&=\int\frac{2u^3-6u^2+6u-2+3u^2-6u+3+4}{u^4}du \\
&=\int\frac{2u^3-3u^2+5}{u^4}du \\
&=\int\left(\frac{2}{u}-\frac{3}{u^2}+\frac{5}{u^4}\right)du \\
&=2\int\frac{1}{u}du-3\int\frac{1}{u^2}du+5\int\frac{1}{u^4}du
\end{align*}
All three are standard integrals:
$$=2\ln|u|+\frac{3}{u}-\frac{5}{3u^3}+C$$
Undoing the substitution(s):
$$=2\ln|x+1|+\frac{3}{x+1}-\frac{5}{3(x+1)^3}+C$$
\subsection*{Problem 70}
$$\int\frac{\sec^2\left(x\right)}{\tan^2\left(x\right)+2\tan\left(x\right)+2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sec^2(x)}{\tan^2(x)+2\tan(x)+1+1}dx \\
&=\int\frac{\sec^2(x)}{(\tan(x)+1)^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=\tan(x)+1 \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{u^2+1}du$$
This integral is standard:
$$=\arctan(u)+C$$
Undoing the substitution(s):
$$=\arctan(\tan(x)+1)+C$$
\subsection*{Problem 71}
$$\int\frac{x^3+x^2+2x+1}{x^4+2x^2+1}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^3+x+x+x^2+1}{(x^2+1)^2}dx \\
&=\int\frac{x+x(x^2+1)+x^2+1}{(x^2+1)^2}dx \\
&=\int\left(\frac{x}{(x^2+1)^2}+\frac{x(x^2+1)}{(x^2+1)^2}+\frac{x^2+1}{(x^2+1)^2}\right)dx \\
&=\int\frac{x}{(x^2+1)^2}dx+\int\frac{x}{x^2+1}dx+\int\frac{1}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+1 \\
du&=2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{u^2}du+\frac{1}{2}\int\frac{1}{u}du$$
Both are standard integrals:
$$=-\frac{1}{2u}+\frac{1}{2}\ln|u|+\arctan(x)+C$$
Undoing the substitution(s):
$$=-\frac{1}{2x^2+2}+\frac{1}{2}\ln\left|x^2+1\right|+\arctan(x)+C$$
\subsection*{Problem 72}
$$\int\frac{3+\cos\left(x\right)}{2-\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=-\int\frac{\cos(x)+3}{\cos(x)-2}dx \\
&=-\int\frac{\cos(x)-2+5}{\cos(x)-2}dx \\
&=-\int\left(\frac{\cos(x)-2}{\cos(x)-2}+\frac{5}{\cos(x)-2}\right)dx \\
&=-\int dx-5\int\frac{1}{\cos(x)-2}dx
\end{align*}
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=-5\int\frac{1}{\frac{1-u^2}{1+u^2}-2}\cdot\frac{2}{1+u^2}du \\
&=-10\int\frac{1}{1-u^2-2(1+u^2)}du \\
&=-10\int\frac{1}{1-u^2-2-2u^2}du \\
&=-10\int\frac{1}{-1-3u^2}du \\
&=10\int\frac{1}{1+3u^2}du
\end{align*}
Both integrals are standard:
$$=-x+\frac{10}{\sqrt{3}}\arctan\left(\sqrt{3}u\right)+C$$
Undoing the substitution(s):
$$=-x+\frac{10}{\sqrt{3}}\arctan\left(\sqrt{3}\tan\left(\frac{x}{2}\right)\right)+C$$
\subsection*{Problem 73}
$$\int x^5\sqrt{x^3-1}dx$$
Rewriting the integral:
$$=\int x^2\cdot x^3\sqrt{x^3-1}dx$$
Substituting
\begin{align*}
u&=x^3-1 \\
u+1&=x^3 \\
du&=3x^2dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{3}\int(u+1)\sqrt{u}du \\
&=\frac{1}{3}\int\left(u^{3/2}+\sqrt{u}\right)du \\
&=\frac{1}{3}\int u^{3/2}du+\frac{1}{3}\int\sqrt{u}du
\end{align*}
Both integrals are standard:
$$=\frac{2}{15}u^{5/2}+\frac{2}{9}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{2}{15}(x^3-1)^{5/2}+\frac{2}{9}(x^3-1)^{3/2}+C$$
\subsection*{Problem 74}
$$\int\frac{1}{2+2\cos\left(x\right)+\sin\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{1}{2+2\left(\frac{1-u^2}{1+u^2}\right)+\frac{2u}{1+u^2}}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{2(1+u^2)+2(1-u^2)+2u}du \\
&=2\int\frac{1}{2+2u^2+2-2u^2+2u}du \\
&=2\int\frac{1}{4+2u}du \\
&=\int\frac{1}{2+u}du
\end{align*}
Substituting
\begin{align*}
v&=2+u \\
dv&=du
\end{align*}
yields:
$$=\int\frac{1}{v}dv$$
This integral is standard:
$$=\ln|v|+C$$
Undoing the substitution(s):
$$=\ln\left|2+\tan\left(\frac{x}{2}\right)\right|+C$$
\subsection*{Problem 75}
$$\int\frac{\sqrt{1+\sin\left(x\right)}}{\sec\left(x\right)}dx$$
Rewriting the integral:
$$=\int\cos(x)\sqrt{1+\sin(x)}dx$$
Substituting
\begin{align*}
u&=1+\sin(x) \\
du&=\cos(x)dx
\end{align*}
yields:
$$=\int\sqrt{u}du$$
This integral is standard:
$$=\frac{2}{3}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{2}{3}(1+\sin(x))^{3/2}+C$$
\subsection*{Problem 76}
$$\int\frac{1}{x^{2/3}\left(1+x^{2/3}\right)}dx$$
Substituting
\begin{align*}
u&=\sqrt[3]{x} \\
du&=\frac{1}{3x^{2/3}}dx
\end{align*}
yields:
$$=3\int\frac{1}{1+u^2}du$$
This integral is standard:
$$=3\arctan(u)+C$$
Undoing the substitution(s):
$$=3\arctan\left(\sqrt[3]{x}\right)+C$$
\subsection*{Problem 77}
$$\int\frac{\sin\left(x\right)}{\sin\left(2x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{\sin(x)}{2\sin(x)\cos(x)}dx \\
&=\frac{1}{2}\int\frac{1}{\cos(x)}dx \\
&=\frac{1}{2}\int\sec(x)dx
\end{align*}
This integral is standard:
$$=\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C$$
\subsection*{Problem 78}
$$\int\sqrt{1+\cos\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1+\cos(x)}\cdot\frac{\sqrt{1-\cos(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sqrt{1-\cos^2(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sqrt{\sin^2(x)}}{\sqrt{1-\cos(x)}}dx \\
&=\int\frac{\sin(x)}{\sqrt{1-\cos(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=1-\cos(x) \\
du&=\sin(x)dx
\end{align*}
yields:
$$=\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=2\sqrt{u}+C$$
Undoing the substitution(s):
$$=2\sqrt{1-\cos(x)}+C$$
\subsection*{Problem 79}
$$\int\sqrt{1+\sin\left(x\right)}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{1+\sin(x)}\cdot\frac{\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{1-\sin^2(x)}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{\sqrt{\cos^2(x)}}{\sqrt{1-\sin(x)}}dx \\
&=\int\frac{\cos(x)}{\sqrt{1-\sin(x)}}dx
\end{align*}
Substituting
\begin{align*}
u&=1-\sin(x) \\
du&=-\cos(x)dx
\end{align*}
yields:
$$=-\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=-2\sqrt{u}+C$$
\subsection*{Problem 80}
$$\int\frac{\sec^2\left(x\right)}{1-\tan^2\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan(x) \\
du&=\sec^2(x)dx
\end{align*}
yields:
$$=\int\frac{1}{1-u^2}du$$
This integral is standard:
$$=\frac{1}{2}\ln\left|\frac{1+u}{1-u}\right|+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|\frac{1+\tan(x)}{1-\tan(x)}\right|+C$$
\subsection*{Problem 81}
$$\int\ln\left(x^2+x+1\right)dx$$
Using integration by parts where
\begin{align*}
u&=\ln(x^2+x+1) &v&=x \\
du&=\frac{2x+1}{x^2+x+1}dx &dv&=dx
\end{align*}
yields:
$$=x\ln(x^2+x+1)-\int\frac{2x^2+x}{x^2+x+1}dx$$
Performing long division on the integral yields:
\begin{align*}
&=-\int\left(2+\frac{-x-2}{x^2+x+1}\right)dx \\
&=-2\int dx+\int\frac{x+2}{x^2+x+1}dx \\
&=-2\int dx+\int\frac{x+\frac{1}{2}+\frac{3}{2}}{x^2+x+1}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int\frac{1}{x^2+x+1}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+\frac{3}{2}\int\frac{1}{x^2+x+\frac{1}{4}+\frac{3}{4}}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+6\int\frac{1}{4x^2+4x+1+3}dx \\
&=-2\int dx+\frac{1}{2}\int\frac{2x+1}{x^2+x+1}dx+6\int\frac{1}{(2x+1)^2+3}dx
\end{align*}
Substituting
\begin{align*}
w&=x^2+x+1 &z&=2x+1 \\
dw&=(2x+1)dx &dz&=2dx
\end{align*}
yields:
$$=-2\int dx+\frac{1}{2}\int\frac{1}{w}dw+3\int\frac{1}{z^2+3}$$
All three integrals are standard:
$$=-2x+\frac{1}{2}\ln|w|+\frac{3}{\sqrt{3}}\arctan\left(\frac{z}{\sqrt{3}}\right)+C$$
Undoing the substitution(s):
$$=x\ln(x^2+x+1)-2x+\frac{1}{2}\ln\left|x^2+x+1\right|+\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+C$$
\subsection*{Problem 82}
$$\int e^x\arcsin\left(e^x\right)dx$$
Substituting
\begin{align*}
u&=e^x \\
du&=e^xdx
\end{align*}
yields:
$$=\int\arcsin(u)du$$
This integral is standard:
$$=u\arcsin(u)+\sqrt{1-u^2}+C$$
Undoing the substitution(s):
$$=e^x\arcsin\left(e^x\right)+\sqrt{1-e^{2x}}+C$$
\subsection*{Problem 83}
$$\int\frac{\arctan\left(x\right)}{x^2}dx$$
Using integration by parts where:
\begin{align*}
u&=\arctan(x) &v&=-\frac{1}{x} \\
du&=\frac{1}{1+x^2}dx &dv&=\frac{1}{x^2}dx
\end{align*}
yields:
$$=-\frac{\arctan(x)}{x}+\int\frac{1}{x(1+x^2)}dx$$
Using partial fraction decomposition
$$=\int\left(\frac{A}{x}+\frac{Bx+C}{1+x^2}\right)dx$$
where $A,B,C\in\mathbb{R}$ yields:
\begin{align*}
&=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx \\
&=\int\frac{1}{x}dx-\int\frac{x}{x^2+1}dx
\end{align*}
Substituting
\begin{align*}
u&=x^2+1 \\
du&=2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=-\frac{1}{2}\ln|u|+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{\arctan(x)}{x}+\ln|x|-\frac{1}{2}\ln\left|x^2+1\right|+C \\
&=-\frac{\arctan(x)}{x}+\ln\left|\frac{x}{\sqrt{x^2+1}}\right|+C
\end{align*}
\subsection*{Problem 84}
$$\int\frac{x^2}{\sqrt{x^2-25}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{x^2}{\sqrt{25\left(\frac{x^2}{25}-1\right)}}dx \\
&=\frac{1}{5}\int\frac{x^2}{\sqrt{\left(\frac{x}{5}\right)^2-1}}dx
\end{align*}
Substituting
\begin{align*}
\sec\theta&=\frac{x}{5} \\
5\sec\theta&=x \\
5\sec\theta\tan\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=\int\frac{(5\sec\theta)^2}{\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta d\theta \\
&=25\int\frac{\sec^3\theta\tan\theta}{\sqrt{\tan^2\theta}}d\theta \\
&=25\int\sec^3\theta d\theta
\end{align*}
This integral is standard:
\begin{align*}
&=25\left(\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C\right) \\
&=\frac{25}{2}\sec\theta\tan\theta+\frac{25}{2}\ln\left|\sec\theta+\tan\theta\right|+C
\end{align*}
Undoing the substitution(s):
\begin{align*}
&=\frac{25}{2}\cdot\frac{x}{5}\cdot\frac{\sqrt{x^2-25}}{5}+\frac{25}{2}\ln\left|\frac{x}{5}+\frac{\sqrt{x^2-25}}{5}\right|+C \\
&=\frac{1}{2}x\sqrt{x^2-25}+\frac{25}{2}\ln\left|x+\sqrt{x^2-25}\right|+C
\end{align*}
\subsection*{Problem 85}
$$\int\frac{x^3}{\left(x^2+1\right)^2}dx$$
Rewriting the integral:
$$=\int\frac{x\cdot x^2}{(x^2+1)^2}dx$$
Substituting
\begin{align*}
u&=x^2+1 \\
u-1&=x^2 \\
du&=2xdx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\frac{u-1}{u^2}du \\
&=\frac{1}{2}\int\left(\frac{1}{u}-\frac{1}{u^2}\right)du \\
&=\frac{1}{2}\int\frac{1}{u}du-\frac{1}{2}\int\frac{1}{u^2}du
\end{align*}
All three integrals are standard:
$$=\frac{1}{2}\ln|u|-\frac{1}{2u}+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\ln\left|x^2+1\right|+\frac{1}{2x^2+2}+C$$
\subsection*{Problem 86}
$$\int\frac{1}{x\sqrt{6x-x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\int\frac{1}{x\sqrt{9-9+6x-x^2}}dx \\
&=\int\frac{1}{x\sqrt{9-(9-6x+x^2)}}dx \\
&=\int\frac{1}{x\sqrt{9-(3-x)^2}}dx \\
&=\int\frac{1}{x\sqrt{9\left(1-\frac{(3-x)^2}{9}\right)}}dx \\
&=\frac{1}{3}\int\frac{1}{x\sqrt{1-\left(\frac{3-x}{3}\right)^2}}dx
\end{align*}
Substituting
\begin{align*}
\sin\theta&=\frac{3-x}{3} \\
3-3\sin\theta&=x \\
-3\cos\theta d\theta&=dx
\end{align*}
yields:
\begin{align*}
&=-\int\frac{\cos\theta}{(3-3\sin\theta)\sqrt{1-\sin^2\theta}}d\theta \\
&=-\int\frac{\cos\theta}{(3-3\sin\theta)\sqrt{\cos^2\theta}}d\theta \\
&=-\int\frac{1}{3-3\sin\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1}{1-\sin\theta}\cdot\frac{1+\sin\theta}{1+\sin\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1+\sin\theta}{1-\sin^2\theta}d\theta \\
&=-\frac{1}{3}\int\frac{1+\sin\theta}{\cos^2\theta}d\theta \\
&=-\frac{1}{3}\int\left(\sec^2\theta+\sec\theta\tan\theta\right)d\theta \\
&=-\frac{1}{3}\int\sec^2\theta d\theta-\frac{1}{3}\int\sec\theta\tan\theta d\theta
\end{align*}
Both integrals are standard:
$$=-\frac{1}{3}\tan\theta-\frac{1}{3}\sec\theta+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{1}{3}\cdot\frac{3-x}{\sqrt{6x-x^2}}-\frac{1}{3}\cdot\frac{3}{\sqrt{6x-x^2}}+C \\
&=\frac{x-3}{3\sqrt{6x-x^2}}-\frac{3}{3\sqrt{6x-x^2}}+C \\
&=\frac{x-6}{3\sqrt{6x-x^2}}+C
\end{align*}
\subsection*{Problem 87}
$$\int\frac{3x+2}{\left(x^2+4\right)^{3/2}}dx$$
Rewriting the integral:
$$\frac{3}{2}\int\frac{2x}{(x^2+4)^{3/2}}dx+\frac{1}{4}\int\frac{1}{\left(\left(\frac{x}{2}\right)^2+1\right)^{3/2}}dx$$
Substituting
\begin{align*}
u&=x^2+4 &\tan\theta&=\frac{x}{2} \\
du&=2xdx &\sec^2\theta d\theta&=\frac{1}{2}dx
\end{align*}
yields:
\begin{align*}
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{\sec^2\theta}{(\tan^2\theta+1)^{3/2}}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{\sec^2\theta}{(\sec^2\theta)^{3/2}}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\frac{1}{\sec\theta}d\theta \\
&=\frac{3}{2}\int\frac{1}{u^{3/2}}du+\frac{1}{2}\int\cos\theta d\theta
\end{align*}
Both integrals are standard:
$$=-\frac{3}{\sqrt{u}}+\frac{1}{2}\sin\theta+C$$
Undoing the substitution(s):
\begin{align*}
&=-\frac{3}{\sqrt{x^2+4}}+\frac{1}{2}\cdot\frac{x}{\sqrt{x^2+4}}+C \\
&=\frac{x-6}{2\sqrt{x^2+4}}+C
\end{align*}
\subsection*{Problem 88}
$$\int x^{3/2}\ln\left(x\right)dx$$
Using integration by parts where
\begin{align*}
u&=\ln(x) &v&=\frac{2}{5}x^{5/2} \\
du&=\frac{1}{x}dx &dv&=x^{3/2}dx
\end{align*}
yields:
$$=\frac{2}{5}x^{5/2}\ln(x)-\frac{2}{5}\int x^{3/2}dx$$
This integral is standard:
$$=\frac{2}{5}x^{5/2}\ln(x)-\frac{4}{25}x^{5/2}+C$$
\subsection*{Problem 89}
$$\int\frac{\sqrt{1+\sin^2\left(x\right)}}{\sec\left(x\right)\csc\left(x\right)}dx$$
Rewriting the integral;
$$=\int\sin(x)\cos(x)\sqrt{1+\sin^2(x)}dx$$
Substituting
\begin{align*}
u&=1+\sin^2(x) \\
du&=2\sin(x)\cos(x)dx
\end{align*}
yields:
$$=\frac{1}{2}\int\sqrt{u}du$$
This integral is standard:
$$=\frac{1}{3}u^{3/2}+C$$
Undoing the substitution(s):
$$=\frac{1}{3}\left(1+\sin^2(x)\right)^{3/2}+C$$
\subsection*{Problem 90}
$$\int\frac{e^{\sqrt{\sin\left(x\right)}}}{\sec\left(x\right)\sqrt{\sin\left(x\right)}}dx$$
Rewriting the integral:
$$=\int\frac{\cos(x)e^{\sqrt{\sin(x)}}}{\sqrt{\sin(x)}}dx$$
Substituting
\begin{align*}
u&=\sqrt{\sin(x)} \\
du&=\frac{\cos(x)}{2\sqrt{\sin(x)}}dx
\end{align*}
yields
$$=2\int e^udu$$
This integral is standard:
$$=2e^u+C$$
Undoing the substitution(s):
$$=2e^{\sqrt{\sin(x)}}+C$$
\subsection*{Problem 91}
$$\int xe^x\sin(x)dx$$
Recall that
$$\int e^x\sin(x)dx=\frac{1}{2}e^x\sin(x)-\frac{1}{2}e^x\cos(x)$$
Using integration by parts where
\begin{center}
\begin{tabular}{ |c|c|c| }
\hline
Sign & D & I \\
\hline
$+$ & $xe^x$ & $\sin(x)$ \\
$-$ & $e^x+xe^x$ & $-\cos(x)$ \\
$+$ & $2e^x+xe^x$ & $-\sin(x)$ \\
\hline
\end{tabular}
\end{center}
yields:
\begin{align*}
&=-xe^x\cos(x)+e^x\sin(x)(1+x)-\int e^x\sin(x)(2+x)dx \\
2\int xe^x\sin(x)dx&=-xe^x\cos(x)+e^x\sin(x)(1+x)-2\int e^x\sin(x)dx \\
&=-xe^x\cos(x)+xe^x\sin(x)+e^x\sin(x)-e^x\sin(x)+e^x\cos(x) \\
\int xe^x\sin(x)dx&=\frac{1}{2}xe^x(\sin(x)-\cos(x))+\frac{1}{2}e^x\cos(x)+C
\end{align*}
\subsection*{Problem 92}
$$\int x^2e^{x^{3/2}}dx$$
Rewriting the integral:
$$=\int x^{3/2}\cdot x^{1/2}e^{x^{3/2}}dx$$
Substituting
\begin{align*}
u&=x^{3/2} \\
du&=\frac{3}{2}x^{1/2}dx
\end{align*}
yields:
$$=\frac{2}{3}\int ue^udu$$
Using integration by parts where
\begin{align*}
w&=u &v&=e^u \\
dw&=du &dv&=e^udu
\end{align*}
yields:
$$=\frac{2}{3}ue^u-\frac{2}{3}\int e^udu$$
This integral is standard:
$$=\frac{2}{3}ue^u-\frac{2}{3}e^u+C$$
Undoing the substitution(s):
$$=\frac{2}{3}x^{3/2}e^{x^{3/2}}-\frac{2}{3}e^{x^{3/2}}+C$$
\subsection*{Problem 93}
$$\int\frac{\arctan\left(x\right)}{\left(x-1\right)^3}dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=-\frac{1}{2(x-1)^2}dx \\
du&=\frac{1}{1+x^2}dx &dv&=\frac{1}{(x-1)^3}dx
\end{align*}
yields:
$$=-\frac{\arctan(x)}{2(x-1)^2}+\frac{1}{2}\int\frac{1}{(1+x^2)(x-1)^2}dx$$
Using partial fraction decomposition
$$=\frac{1}{2}\int\left(\frac{Ax+B}{1+x^2}+\frac{C}{(x-1)^2}+\frac{D}{x-1}\right)dx$$
where $A,B,C,D\in\mathbb{R}$ yields:
\begin{align*}
&=\frac{1}{2}\bigintss\left(\frac{\frac{1}{2}x+0}{x^2+1}+\frac{-\frac{1}{2}}{x-1}+\frac{\frac{1}{2}}{\left(x-1\right)^2}\right)dx \\
&=\frac{1}{4}\int\left(\frac{x}{x^2+1}-\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)dx \\
&=\frac{1}{4}\int\frac{x}{x^2+1}dx-\frac{1}{4}\int\frac{1}{x-1}dx+\frac{1}{4}\int\frac{1}{(x-1)^2}dx
\end{align*}
Substituting
\begin{align*}
w_1&=x^2+1 &w_2&=x-1 \\
dw_1&=2xdx &dw_2&=dx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{8}\int\frac{1}{w_1}dw_1-\frac{1}{4}\int\frac{1}{w_2}dw_2+\frac{1}{4}\int\frac{1}{w_2^2}dw_2
\end{align*}
All three integrals are standard:
\begin{align*}
&=\frac{1}{8}\ln\left|w_1\right|-\frac{1}{4}\ln\left|w_2\right|-\frac{1}{4w_2} \\
&=\frac{1}{8}\ln\left|\frac{w_1}{w_2^2}\right|-\frac{1}{4w_2}
\end{align*}
Undoing the substitution(s):
$$=-\frac{\arctan(x)}{2(x-1)^2}+\frac{1}{8}\ln\left|\frac{x^2+1}{(x-1)^2}\right|-\frac{1}{4(x-1)}+C$$
\subsection*{Problem 94}
$$\int\ln\left(1+\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=1+\sqrt{x} \\
u-1&=\sqrt{x} \\
u^2-2u+1&=x \\
(2u-2)du&=dx
\end{align*}
yields:
\begin{align*}
&=\int\ln(u)(2u-2)du \\
&=2\int u\ln(u)du-2\int\ln(u)du
\end{align*}
Using integration by parts where
\begin{align*}
w&=\ln(u) &v&=\frac{1}{2}u^2 \\
dw&=\frac{1}{u}du &dv&=udu
\end{align*}
yields:
\begin{align*}
&=2\left(\frac{1}{2}u^2\ln(u)-\frac{1}{2}\int udu\right) \\
&=u^2\ln(u)-\int udu
\end{align*}
The remaining two integrals are standard:
\begin{align*}
&=u^2\ln(u)-\frac{1}{2}u^2-2u\ln(u)+2u+C \\
&=u\ln(u)\left(u-2\right)-\frac{1}{2}u^2+2u+C
\end{align*}
Undoing the substitution(s):
$$=\left(1+\sqrt{x}\right)\ln\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)-\frac{1}{2}\left(1+\sqrt{x}\right)^2+2\sqrt{x}+C$$
\subsection*{Problem 95}
$$\int\frac{2x+3}{\sqrt{3+6x-9x^2}}dx$$
Rewriting the integral:
\begin{align*}
&=\bigintss\frac{2x-\frac{2}{3}+\frac{11}{3}}{\sqrt{3+6x-9x^2}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-1+6x-9x^2}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-(1-6x+9x^2)}}dx \\
&=-\frac{1}{9}\int\frac{6-18x}{\sqrt{3+6x-9x^2}}+\frac{11}{3}\int\frac{1}{\sqrt{4-(1-3x)^2}}dx
\end{align*}
Substituting
\begin{align*}
u&=3+6x-9x^2 &v&=1-3x \\
du&=(6-18x)dx &dv&=-3dx
\end{align*}
yields:
$$=-\frac{1}{9}\int\frac{1}{\sqrt{u}}du-\frac{11}{9}\int\frac{1}{\sqrt{4-v^2}}dv$$
Both integrals are standard:
$$=-\frac{2}{9}\sqrt{u}-\frac{11}{9}\arcsin\left(\frac{v}{2}\right)+C$$
Undoing the substitution(s):
$$=-\frac{2}{9}\sqrt{3+6x-9x^2}-\frac{11}{9}\arcsin\left(\frac{1-3x}{2}\right)+C$$
\subsection*{Problem 96}
$$\int\frac{1}{2+2\sin\left(x\right)+\cos\left(x\right)}dx$$
Substituting
\begin{align*}
u&=\tan\left(\frac{x}{2}\right) \\
du&=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx \\
\frac{2}{1+u^2}du&=dx
\end{align*}
yields:
\begin{align*}
&=\bigintss\frac{1}{2+2\left(\frac{2u}{1+u^2}\right)+\left(\frac{1-u^2}{1+u^2}\right)}\cdot\frac{2}{1+u^2}du \\
&=2\int\frac{1}{2+2u^2+4u+1-u^2}du \\
&=2\int\frac{1}{u^2+4u+3}du \\
&=2\int\frac{1}{(u+3)(u+1)}du \\
&=2\bigintss\left(\frac{-\frac{1}{2}}{u+3}+\frac{\frac{1}{2}}{u+1}\right)du \\
&=-\int\frac{1}{u+3}du+\int\frac{1}{u+1}du
\end{align*}
Substituting
\begin{align*}
v_1&=u+3 &v_2&=u+1 \\
dv_1&=du &dv_2&=du
\end{align*}
yields:
$$=-\int\frac{1}{v_1}dv_1+\int\frac{1}{v_2}dv_2$$
Both integrals are standard:
\begin{align*}
&=-\ln|v_1|+\ln|v_2|+C \\
&=\ln\left|\frac{v_2}{v_1}\right|+C
\end{align*}
Undoing the substitution(s):
$$\ln\left|\frac{\tan\left(\frac{x}{2}\right)+1}{\tan\left(\frac{x}{2}\right)+3}\right|+C$$
\subsection*{Problem 97}
$$\int\frac{\sin^3\left(x\right)}{\cos\left(x\right)-1}dx$$
Rewriting the integral:
\begin{align*}
&=-\int\frac{\sin(x)\cdot\sin^2(x)}{1-\cos(x)}dx \\
&=-\int\frac{\sin(x)(1-\cos^2(x))}{1-\cos(x)}dx \\
&=-\int\frac{\sin(x)(1-\cos(x))(1+\cos(x))}{1-\cos(x)}dx \\
&=-\int\sin(x)(1+\cos(x))dx
\end{align*}
Substituting
\begin{align*}
u&=1+\cos(x)\\
du&=-\sin(x)dx
\end{align*}
yields:
$$=\int udu$$
This integral is standard:
$$=\frac{1}{2}u^2+C$$
Undoing the substitution(s):
$$=\frac{1}{2}(1+\cos(x))^2+C$$
\subsection*{Problem 98}
$$\int x^{3/2}\arctan\left(\sqrt{x}\right)dx$$
Using integration by parts where:
\begin{align*}
u&=\arctan\left(\sqrt{x}\right) &v&=\frac{2}{5}x^{5/2} \\
dw&=\frac{1}{2\sqrt{x}(1+x)}dx &dv&=x^{3/2}dx
\end{align*}
yields:
\begin{align*}
&=\frac{2}{5}x^{5/2}\arctan\left(\sqrt{x}\right)-\frac{1}{5}\int\frac{x^2}{1+x}dx
\end{align*}
Substituting
\begin{align*}
w&=1+x \\
w-1&=x \\
dw&=dx
\end{align*}
yields:
\begin{align*}
&=-\frac{1}{5}\int\frac{(w-1)^2}{w}dw \\
&=-\frac{1}{5}\int\frac{w^2-2w+1}{w}dw \\
&=-\frac{1}{5}\int\left(w-2+\frac{1}{w}\right)dw \\
&=-\frac{1}{5}\int wdw+\frac{2}{5}\int dw-\frac{1}{5}\int\frac{1}{w}dw
\end{align*}
All three integrals are standard:
$$-\frac{1}{10}w^2+\frac{2}{5}w-\frac{1}{5}\ln|w|+C$$
Undoing the substitution(s):
$$=\frac{2}{5}x^{5/2}\arctan\left(\sqrt{x}\right)-\frac{1}{10}(1+x)^2+\frac{2}{5}(1+x)-\frac{1}{5}\ln|1+x|+C$$
\subsection*{Problem 99}
$$\int\arcsec\left(\sqrt{x}\right)dx$$
Substituting
\begin{align*}
u&=\sqrt{x} \\
u^2&=x \\
2udu&=dx
\end{align*}
yields:
$$=2\int u\arcsec(u)du$$
Using integration by parts where:
\begin{align*}
w&=\arcsec(u) &v&=\frac{1}{2}u^2 \\
dw&=\frac{1}{u\sqrt{u^2-1}}du &dv&=udu
\end{align*}
yields:
\begin{align*}
&=2\left(\frac{1}{2}u^2\arcsec(u)-\frac{1}{2}\int\frac{u^2}{u\sqrt{u^2-1}}du\right) \\
&=u^2\arcsec(u)-\int\frac{u}{\sqrt{u^2-1}}du
\end{align*}
Substituting
\begin{align*}
z&=u^2-1 \\
dz&=2udu
\end{align*}
yields:
$$=u^2\arcsec(u)-\frac{1}{2}\int\frac{1}{\sqrt{z}}dz$$
This integral is standard:
$$=u^2\arcsec(u)-\sqrt{z}+C$$
Undoing the substitution(s):
\begin{align*}
&=x\arcsec\left(\sqrt{x}\right)-\sqrt{x-1}+C
\end{align*}
\subsection*{Problem 100}
$$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx$$
Substituting
\begin{align*}
\sin\theta&=x^2 \\
\cos\theta d\theta&=2xdx
\end{align*}
yields:
\begin{align*}
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{1-\sin\theta}{1+\sin\theta}\cdot\frac{1-\sin\theta}{1-\sin\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}d\theta \\
&=\frac{1}{2}\int\cos\theta\cdot\frac{1-\sin\theta}{\cos\theta}d\theta \\
&=\frac{1}{2}\int(1-\sin\theta)d\theta \\
&=\frac{1}{2}\int d\theta-\frac{1}{2}\int\sin\theta d\theta
\end{align*}
Both integrals are standard:
$$=\frac{1}{2}\theta+\frac{1}{2}\cos\theta+C$$
Undoing the substitution(s):
$$=\frac{1}{2}\arcsin\left(x^2\right)+\frac{1}{2}\sqrt{1-x^4}+C$$
\newpage
\section{Advanced Integrals - Proofs}
\subsection*{Advanced Integral 1}
$$\int\sec^3(x)dx$$
Rewriting the integral:
$$=\int\sec(x)\sec^2(x)dx$$
Using integration by parts where
\begin{align*}
u&=\sec(x) &v&=\tan(x) \\
du&=\sec(x)\tan(x)dx &dv&=\sec^2(x)dx
\end{align*}
yields:
\begin{align*}
&=\sec(x)\tan(x)-\int\sec(x)\tan^2(x)dx \\
&=\sec(x)\tan(x)-\int\sec(x)(\sec^2(x)-1)dx \\
&=\sec(x)\tan(x)-\int(\sec^3(x)-\sec(x))dx \\
&=\sec(x)\tan(x)-\int\sec^3(x)dx+\int\sec(x)dx \\
2\int\sec^3(x)dx&=\sec(x)\tan(x)+\ln\left|\sec(x)+\tan(x)\right|+C \\
&=\frac{1}{2}\sec(x)\tan(x)+\frac{1}{2}\ln\left|\sec(x)+\tan(x)\right|+C
\end{align*}
\subsection*{Advanced Integral 2}
$$\int\arcsin(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arcsin(x) &v&=x \\
du&=\frac{1}{\sqrt{1-x^2}}dx &dv&=dx
\end{align*}
yields:
$$x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=1-x^2 \\
du&=-2xdx
\end{align*}
yields:
$$=\frac{1}{2}\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=\sqrt{u}+C$$
Undoing the substitution(s):
$$=x\arcsin(x)+\sqrt{1-x^2}+C$$
\subsection*{Advanced Integral 3}
$$\int\arccos(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arccos(x) &v&=x \\
du&=-\frac{1}{\sqrt{1-x^2}}dx &dv&=dx
\end{align*}
yields:
$$x\arccos(x)+\int\frac{x}{\sqrt{1-x^2}}dx$$
Substituting
\begin{align*}
u&=1-x^2 \\
du&=-2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{\sqrt{u}}du$$
This integral is standard:
$$=-\sqrt{u}+C$$
Undoing the substitution(s):
$$=x\arccos(x)-\sqrt{1-x^2}+C$$
\subsection*{Advanced Integral 4}
$$\int\arctan(x)dx$$
Using integration by parts where
\begin{align*}
u&=\arctan(x) &v&=x \\
du&=\frac{1}{1+x^2}dx &dv&=dx
\end{align*}
yields:
$$=x\arctan(x)-\int\frac{x}{1+x^2}dx$$
Substituting
\begin{align*}
u&=1+x^2 \\
du&=2xdx
\end{align*}
yields:
$$=-\frac{1}{2}\int\frac{1}{u}du$$
This integral is standard:
$$=x\arctan(x)-\frac{1}{2}\ln\left|1+x^2\right|+C$$
\subsection*{Advanced Integral 5}
$$\int\sqrt{x^2+a^2}dx$$
Rewriting the integral:
\begin{align*}
&=\int\sqrt{a^2\left(\frac{x^2}{a^2}+1\right)}dx \\
&=a\int\sqrt{\left(\frac{x}{a}\right)^2+1}dx
\end{align*}
Substituting
$$$$
\subsection*{Advanced Integral 6}
$$\int\sqrt{x^2-a^2}dx$$
\subsection*{Advanced Integral 7}
$$\int\sqrt{a^2-x^2}dx$$
\subsection*{Advanced Integral 8}
$$\int\frac{1}{x^2+a^2}dx$$
\subsection*{Advanced Integral 9}
$$\int\frac{1}{x^2-a^2}dx$$
\subsection*{Advanced Integral 10}
$$\int\frac{1}{a^2-x^2}dx$$
\subsection*{Advanced Integral 11}
$$\int\frac{1}{\sqrt{x^2+a^2}}dx$$
\subsection*{Advanced Integral 12}
$$\int\frac{1}{\sqrt{x^2-a^2}}dx$$
\subsection*{Advanced Integral 13}
$$\int\frac{1}{\sqrt{a^2-x^2}}dx$$
\subsection*{Advanced Integral 14}
$$\int\sin^n(x)dx$$
\subsection*{Advanced Integral 15}
$$\int\cos^n(x)dx$$
\end{document}