Answer
$-\sin^{-1}(e^{-x})+C$
Work Step by Step
$I=\displaystyle \int\frac{e^{-x}dx}{\sqrt{1-(e^{-x})^{2}}}=\quad\left[\begin{array}{l}
u=e^{-x}\\
du=-e^{-x}dx
\end{array}\right]$
$=\displaystyle \int\frac{-du}{\sqrt{1-u^{2}}}$
$=-\sin^{-1}u+C$
... bring back x
$=-\sin^{-1}(e^{-x})+C$
