Answer
$2\tan^{-1}\sqrt{e^{x}-1}+C$
Work Step by Step
$\displaystyle \int\frac{dx}{\sqrt{e^{x}-1}}=$
$\left[\begin{array}{ll}
u=\sqrt{e^{x}-1} & u^{2}=e^{x}-1\\
& 2udu=e^{x}dx\\
& \frac{2u}{u^{2}+1}du=dx
\end{array}\right], \quad e^{x}=u^{2}+1$
$=\displaystyle \int\frac{1}{u}\cdot\frac{2u}{u^{2}+1}du$
$=2\displaystyle \int\frac{1}{u^{2}+1}du$
... a basic integral,
$=2\tan^{-1}u+C$
$=2\tan^{-1}\sqrt{e^{x}-1}+C$
