Answer
$x-\ln\left(1+\sqrt{1-e^{2x}}\right) +C$
Work Step by Step
$\displaystyle \int\frac{dx}{\sqrt{1-e^{2x}}}=\quad \left[\begin{array}{ll}
u=e^{x} & \\
du=e^{x}dx & dx=\frac{du}{e^{x}}=\frac{du}{u}
\end{array}\right]$
$=\displaystyle \int\frac{du}{u\sqrt{1^2-u^{2}}}=$
Table of integrals:
$\color{blue}{35. \quad\displaystyle \int\frac{du}{u\sqrt{\mathrm{a}^{2}-u^{2}}}=-\frac{1}{a}\ln\left|\frac{a+\sqrt{a^{2}-u^{2}}}{u}\right| +C }$
$=-\displaystyle \frac{1}{1}\ln\left|\frac{1+\sqrt{1^{2}-u^{2}}}{u}\right| +C$
... bring back x...
$=-\ln\left|\dfrac{1+\sqrt{1-e^{2x}}}{e^{x}}\right| +C$
$=-(\ln\left|1+\sqrt{1-e^{2x}}\right|-x) +C$
... the argument of ln is always positive....
$=x-\ln\left(1+\sqrt{1-e^{2x}}\right) +C$
