Answer
$$2\tan^{-1}(\sqrt{e^x-1})+c$$
Work Step by Step
Given $$\int \frac{dx}{\sqrt{e^x-1}}$$
Let $u^2= e^x-1\ \ \Rightarrow \ 2udu= e^xdx\ \ \Rightarrow dx=\dfrac{ 2udu}{u^2+1}$, then
\begin{align*}
\int \frac{dx}{\sqrt{e^x-1}}&=\int \frac{2udu}{u(u^2+1)}\\
&=\int \frac{2 du}{ u^2+1 }\\
&=2\tan^{-1}u+c\\
&=2\tan^{-1}(\sqrt{e^x-1})+c
\end{align*}
