Answer
$$\int \frac{dx}{\sqrt{e^{2x}-1}}
=\cos^{-1}e^{-x}+c.$$
Work Step by Step
Since $ u=e^{-x}$, then $ du=- e^{-x}dx $ and hence
$$\int \frac{dx}{\sqrt{e^{2x}-1}}=-\int \frac{du}{u\sqrt{(1/u^2)-1}} =-\int \frac{du}{
\sqrt{1-u^2}} \\
=\cos^{-1 }u+c=\cos^{-1}e^{-x}+c.$$
(We could also have integrated to inverse sine by excluding the negative sign.)
