Answer
\[\frac{1}{2}\sin^{-1}(e^{2x})+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int\frac{e^{2x}}{\sqrt {1-e^{4x}}}dx\]
\[I=\frac{1}{2}\int\frac{2e^{2x}}{\sqrt {1-(e^{2x})^2}}dx\]
Put \[t=e^{2x}\;\;\Rightarrow \;\;dt=2e^{2x}\]
\[I=\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}}\]
\[I=\frac{1}{2}\sin^{-1}(t)+C\]
Where $C$ is constant of integration
\[\Rightarrow I=\frac{1}{2}\sin^{-1}(e^{2x})+C\]
Hence , \[I=\frac{1}{2}\sin^{-1}(e^{2x})+C\]
