Answer
$$\ln \left|e^{x}-1\right|-\ln e^{x}+C$$
Work Step by Step
Given
$$ \int \frac{e^{x} d x}{e^{2 x}-e^{x}}$$
Let
$$u=e^x \ \ \ \ \ \to \ \ \ \ du=e^x dx $$
Then
\begin{aligned}
\int \frac{e^{x} d x}{e^{2 x}-e^{x}}&=\int \frac{u \cdot \frac{1}{u} d u}{u^{2}-u}\\
&=\int \frac{1}{u(u-1)} d u
\end{aligned}
Since
\begin{align*}
\int \frac{1}{u(u-1)} &=\frac{A}{u}+\frac{B}{u-1}\\
&=\frac{(A+B) u-A}{u(u-1)}\\
1&= (A+B) u-A
\end{align*}
Then
\begin{align*}
\text{at } u&=0 \ \ \ \ \to A=-1 \\
\text{at } u&= 1\ \ \ \ \to B=1
\end{align*}
Hence
\begin{align*}
\int \frac{e^{x} d x}{e^{2 x}-e^{x}}&=-\int \frac{1}{u} d u+\int \frac{1}{u-1} d u\\
&=\ln |u-1|-\ln |u|+C\\
&=\ln \left|e^{x}-1\right|-\ln e^{x}+C
\end{align*}
