Answer
$\displaystyle \frac{1}{2}\tan^{-1}e^{2x}+C$
Work Step by Step
$\displaystyle \int\frac{e^{2x}}{1+e^{4x}}dx=\left[\begin{array}{l}
u=e^{2x}\\
du=2e^{2x}dx
\end{array}\right]$
$=\displaystyle \frac{1}{2}\int\frac{du}{1+u^{2}}$
$=\displaystyle \frac{1}{2}\tan^{-1}u+C$
$=\displaystyle \frac{1}{2}\tan^{-1}e^{2x}+C$
