Answer
$\sqrt{e^{2x}-1}-\cos^{-1}(e^{-x})+C$.
Work Step by Step
$\displaystyle \int\sqrt{e^{2x}-1}dx=\quad \left[\begin{array}{ll}
u=e^{x} & \\
du=e^{x}dx & dx=\frac{du}{e^{x}}=\frac{du}{u}
\end{array}\right]$
$=\displaystyle \int\frac{\sqrt{u^{2}-1^{2}}}{u}du$
Table of integrals:
$\color{blue}{41.\quad \displaystyle \int\frac{\sqrt{u^{2}-a^{2}}}{u}du=\sqrt{u^{2}-a^{2}}-\mathrm{a}\cos^{-1}\frac{a}{|u|}+C }$
$=\displaystyle \sqrt{u^{2}-1}-\cos^{-1}(\frac{1}{u})+C$
... bring back x...
$=\sqrt{e^{2x}-1}-\cos^{-1}(e^{-x})+C$.
