Answer
$$
\int \frac{e^{x}}{3-e^{2 x}} d x=\frac{1}{2 \sqrt{3}} \ln \left|\frac{e^{x}+\sqrt{3}}{e^{x}-\sqrt{3}}\right|+C
$$
Work Step by Step
$$
\int \frac{e^{x}}{3-e^{2 x}} d x
$$
If we make the substitution
$$
u=u=e^{x}, \quad d u=e^{x} d x
$$
and we look at the Table of Integrals , we see that the closest entry is number $101 $ with $a=\sqrt{3}$ so we have:
\begin{aligned}
\int \frac{e^{x}}{3-e^{2 x}} d x &=\int \frac{d u}{a^{2}-u^{2}} \\
&\stackrel{19}{=} \frac{1}{2 a} \ln \left|\frac{u+a}{u-a}\right|+C\\
&=\frac{1}{2 \sqrt{3}} \ln \left|\frac{e^{x}+\sqrt{3}}{e^{x}-\sqrt{3}}\right|+C
\end{aligned}
