Answer
$$\int \frac{d y}{ \sqrt{e ^{2y}-1}}=\sec ^{-1}\left|e^{y}\right|+C $$
Work Step by Step
Given $$\int \frac{d y}{ \sqrt{e ^{2y}-1}}$$
Let $$ u=e^y \Rightarrow y=\ln u\Rightarrow d y=\frac{d u}{u}$$
So, we have
\begin{aligned}
I&= \int \frac{d u}{u \sqrt{u^{2}-1}}\\
&=\sec ^{-1}|u|+C\\
&=\sec ^{-1}\left|e^{y}\right|+C
\end{aligned}
