Answer
$$\frac{1}{2}\tan^{-1}(e^{2x})+c$$
Work Step by Step
Given
$$\int\frac{e^{2x}dx}{1+e^{4x}}$$
Let $u=e^{2x} \ \ \Rightarrow du=2e^{2x}dx $, then
\begin{align*}
\int\frac{e^{2x}dx}{1+e^{4x}}&=\frac{1}{2}\int\frac{ du}{1+u^{2}}\\
&=\frac{1}{2}\tan^{-1}u+c\\
&=\frac{1}{2}\tan^{-1}(e^{2x})+c
\end{align*}
