Answer
$\displaystyle \frac{1}{2\sqrt{3}}\ln|\frac{e^{x}+\sqrt{3}}{e^{x}-\sqrt{3}}|+C$
Work Step by Step
$\displaystyle \int\frac{e^{x}}{3-e^{2x}}dx=\quad $substitute $\left[\begin{array}{l}
u=e^{x}\\
du=e^{x}dx
\end{array}\right]$
$=\displaystyle \int\frac{du}{(\sqrt{3})^{2}-u^{2}}=$
Table of integrals:
$\color{blue}{19. \quad \displaystyle \int\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\ln \left|\frac{u+a}{u-a}\right|+C }$
=$\displaystyle \frac{1}{2\sqrt{3}}\ln|\frac{u+\sqrt{3}}{u-\sqrt{3}}|+C$
...bring x back ($u=e^{x}$)...
$=\displaystyle \frac{1}{2\sqrt{3}}\ln|\frac{e^{x}+\sqrt{3}}{e^{x}-\sqrt{3}}|+C$
