Answer
$\frac{12-3\pi }{2}$
Work Step by Step
Given $$\int_0^{\ln 10} \frac{e^x \sqrt{e^x-1}}{e^x+8} d x $$
Let $$u^2=e^x-1\ \ \to \ \ 2 u d u=e^x d x$$
At $x=0\ \ \to \ u= 0,\ \ \ x=\ln (10)\to \ \ u =3$, then
\begin{aligned}
\int_0^{\ln 10} \frac{e^x \sqrt{e^x-1}}{e^x+8} d x & =\int_0^3 \frac{u \cdot 2 u d u}{u^2+9}\\
&=2 \int_0^3 \frac{u^2}{u^2+9} d u\\
&=2 \int_0^3\left(1-\frac{9}{u^2+9}\right) d u \\
& =2\left[u-\frac{9}{3} \tan ^{-1}\left(\frac{u}{3}\right)\right]_0^3\\
&=2\left[\left(3-3 \tan ^{-1} 1\right)-0\right]\\
&=2\left(3-3 \cdot \frac{\pi}{4}\right)\\
&=6-\frac{3 \pi}{2}\\
&=\frac{12-3\pi }{2}
\end{aligned}
