Problem source

\begin{questionparts}
\item
Use the substitution $u^2=2x+1$ to show that, for $x>4$, 
\[
\int \frac{3} { ( x-4 ) \sqrt {2x+1}} \; 
\d x = \ln \l \frac{\sqrt{2x+1}-3} {\sqrt{2x+1}+3} \r + K\,,
\]
where $K$ is a constant.
 
\item Show that  $ \displaystyle \int_{\ln 3}^{\ln 8}
 \frac{2} { \e^x \sqrt{ \e^x + 1}}\; \mathrm{d}x\, = 
 \frac 7{12} + \ln \frac23 $
.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& I &= \int \frac{3}{(x-4)\sqrt{2x+1}}\, \d x \\
u^2 =2x+1, 2u \frac{\d u}{\d x}=2: && &= \int \frac{3}{\left(\frac{u^2-1}{2}-4\right)u} u \d u \\
&&&= \int \frac{6}{u^2-9} \d u \\
&&&= \int \frac{6}{(u-3)(u+3)} \d u\\
&&&= \int \left ( \frac{1}{u-3} - \frac{1}{u+3} \right )\d u \\
&&&= \ln (u-3) - \ln (u+3) + C \\
&&&= \ln \frac{u-3}{u+3} + C \\
&&&= \ln \left (\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3} \right) + C
\end{align*}

\item \begin{align*}
&& I &= \int_{\ln 3}^{\ln 8} \frac{2}{e^x\sqrt{e^x+1}} \d x \\
u = e^x, \frac{\d u}{\d x} = e^x: &&&= \int_{u=3}^{u=8} \frac{2}{u\sqrt{u+1}} \frac{1}{u} \d u \\
v^2=u+1, 2v \frac{\d v}{\d u} = 1: &&&= \int_{v=2}^{v=3} \frac{2}{v(v^2-1)^2} \d v \\
&&&= \int_2^3 \left ( \frac{2}{v} - \frac{}{v-1} - \frac{1}{2(v-1)^2} - \frac{1}{v+1} - \frac{1}{2(v+1)^2}\right) \d v \\
&&&= \left [2\ln v - \ln(v^2-1)+\frac12(v-1)^{-1}+\frac12(v+1)^{-1} \right]_2^3 \\
&&&= \left ( 2\ln3-\ln 8+\frac14+\frac18\right)-\left ( 2\ln2-\ln 3+\frac12+\frac16\right) \\
&&&= 

\end{align*}
\end{questionparts}