## Calculus (3rd Edition)

$$\int \frac{dx}{\sqrt{e^{2x}-1}} =\cos^{-1}e^{-x}+c.$$
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.)