Answer
$ 6-\displaystyle \frac{3\pi}{2}$
Work Step by Step
$\displaystyle \int\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8}dx=\quad\left[\begin{array}{lll}
u=\sqrt{e^{x}-1}, & u^{2}=e^{x}-1 & u^{2}+9=e^{x}+8\\
& 2udu=e^{x}dx & \\
& \frac{2udu}{u^{2}+1}=dx &
\end{array}\right]$
$=\displaystyle \int\frac{(u^{2}+1)u2u}{(u^{2}+9)(u^{2}+1)}du=\int\frac{2u^{2}}{u^{2}+9}du$
$=2\displaystyle \int\frac{u^{2}+9-9}{u^{2}+9}du$
$=2\displaystyle \int 1du-2\int\frac{9}{u^{2}+3^{2}}du$
$=2u-18\displaystyle \cdot\frac{1}{3}\tan^{-1}(\frac{u}{3})+C$
$\displaystyle \int_{0}^{\ln 10}\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8}dx=\qquad \left[\begin{array}{l}
x=\ln 10\Rightarrow u=\sqrt{9}=3\\
x=0\Rightarrow u=\sqrt{e^{o}-1}=0
\end{array}\right]$
$=\left[2u-6\tan^{-1}(\frac{u}{3})\right]_{0}^{3}$
$=6-6\tan^{-1}1-(0)$
$=6-\displaystyle \frac{6\pi}{4}$
$=6-\displaystyle \frac{3\pi}{2}$
