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\problemgrouptitle{Integration Methods Review}
  
\begin{problemblock}[2]{Evaluate the given integral.}

\item $\displaystyle\int x\, \tan^{-1}\! x\, dx$
\answer{
$\displaystyle \frac 12 x^2 \tan^{-1}\! x-\frac 12 x+\frac 12 \tan^{-1}\! x+C$
}

\item $\displaystyle\int \tan x\, \sec^3 x\, dx$
\answer{
$\displaystyle \frac 13 \sec^3 x+C$
}

\item $\displaystyle\int_0^2  \ln (2+x)\, dx$
\answer{
$\displaystyle 6\ln 2-2$
}

\item $\displaystyle\int \frac 1 {\left (x^2+36\right)^{5/2}}\, dx$
\answer{
$\displaystyle \frac 1{1296}\, \left[ \frac x{\sqrt{x^2+36}}-\frac 13 \, \left ( \frac x{\sqrt{x^2+36}}  \right)^3  \right]+C$
}

\item $\displaystyle\int \frac {x-2} {\left (x+1\right)^{5}}\, dx$
\answer{
$\displaystyle -\frac 13\, \frac 1{(x+1)^3}+\frac 34\, \frac 1{(x+1)^4}+C$
}

\item $\displaystyle\int \frac 4 {x^3+2x}\, dx$
\answer{
$\displaystyle 2\ln |x|-\ln |x^2+2|+C$
}

\item $\displaystyle\int \frac 1 {\sqrt{x^2+6x+12}}\, dx$
\answer{
$\displaystyle \, \ln \left | \frac{\sqrt{x^2+6x+12}+x+3}{\sqrt 3} \right|+C$
}

\item $\displaystyle\int_0^{\ln 2} \frac {e^{3x}} {1+e^x}\, dx$
\answer{
$\displaystyle \, \frac {1}2+\ln \frac 32$
}

\item $\displaystyle\int \frac {4x^2-8x-6} {(x-1)(x-2)(x-3)}\, dx$
\answer{
$\displaystyle -5\ln |x-1|+6\ln |x-2|+3\ln |x-3|+C$
}

\item $\displaystyle\int \frac {6x^3-4x^2+5} {x^4+5x^2+4}\, dx$
\answer{
$\displaystyle -\ln |x^2+1|+3\tan^{-1}\! x+4\ln |x^2+4|-\frac 72 \tan^{-1}\!\left( \frac x2 \right)+C$
}

\item $\displaystyle\int x^{5/2}\, \ln x\, dx$
\answer{
$\displaystyle \frac 27\, x^{7/2}\ln x-\frac 4{49}\, x^{7/2}+C$
}

\item $\displaystyle\int \frac {\cos x} {\sqrt{1+\sin^2 x}}\, dx$
\answer{
$\displaystyle \ln |\sqrt{1+\sin ^2 x}+\sin x|+C$
}

\end{problemblock}

\begin{problemblock}[2]{Determine whether the given integral is convergent or divergent.}

\item $\displaystyle\int_0^{\infty}x\, e^{-3x}\, dx$
\answer{
convergent
}

\item $\displaystyle\int_0^{\pi}\frac {\sin x}{\sqrt{1+\cos x}}\, dx$
\answer{
convergent
}

\item $\displaystyle\int_{-\infty}^{\infty} \frac 1{5+x^2}\, dx$
\answer{
convergent
}

\end{problemblock}

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