Answer
$ \frac{1}{2}\tan^{-1}(e^x)+c,c \in \mathbb R$
Work Step by Step
$\int \frac{e^{x}}{1-e^{2x}}dx=\int \frac{e^{x}}{1-(e^{x})^{2} }dx$
Take $t=e^{x}$ so $dt=(e^{x})'dx \to dt=e^{x}dx$
$\int \frac{e^{x}}{1-(e^{x})^{2} }dx \to \int \frac{1}{1-t^{2} }dt $
Using the partial decomposition of fractions it follows:
$ \int \frac{1}{1-t^{2} }dt \to \int \frac{1}{(1-t)(1+t) }dt \to \int \frac{1}{2}(\frac{1}{1-t }+\frac{1}{1+t })dt \to \int \frac{1}{2}(\frac{-(1-t)'}{1-t }+\frac{(1+t)'}{1+t })dt \to \int \frac{1}{2}\frac{-(1-t)'}{1-t }dt+\int \frac{1}{2}\frac{(1+t)'}{1+t } dt $
Using the fundamental theorem of calculus it follows:
$\int \frac{1}{2}\frac{-(1-t)'}{1-t }dt+\int \frac{1}{2}\frac{(1+t)'}{1+t } dt \to -\frac{1}{2}\ln(|1-t|)+\frac{1}{2}\ln(|1+t|)+c $
So substitute back $t$ by $e^{x}$ it follows:
$ -\frac{1}{2}\ln(|1-e^{x}|)+\frac{1}{2}\ln(|1+e^{x}|)+c=\frac{1}{2}\tanh^{-1}(e^x)+c,c \in \mathbb R$
