Answer
$ -\sin^{-1}(e^{-x})+C$
Work Step by Step
Given
$$\int \frac{d x}{e^x \sqrt{1-e^{-2 x}}}$$
Let
$$u= e^{-x}\ \ \ \to\ \ \ du =\frac{-dx}{e^x}$$
Then
\begin{aligned}
\int \frac{d x}{e^x \sqrt{1-e^{-2 x}}}&= \int \frac{-d u}{ \sqrt{1-u^2}}\\
&=-\sin^{-1}(u)+C\\
&= -\sin^{-1}(e^{-x})+C
\end{aligned}
