Answer
\[\frac{1}{2}\ln \,\left( {{e^{2x}} + 5} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^{2x}}}}{{{e^{2x}} + 5}}} dx \hfill \\
Let\,\,u = {e^{2x}} + 5\,\,,\,\,So\,\,that \hfill \\
du = 2{e^{2x}}dx \hfill \\
Then \hfill \\
\frac{1}{2}\int_{}^{} {\frac{{2{e^{2x}}}}{{{e^{2x}} + 5}}dx} = \frac{1}{2}\int_{}^{} {\frac{{du}}{u}} \hfill \\
Integrating \hfill \\
\frac{1}{2}\ln \left| u \right| + C \hfill \\
Substituting\,\,u = {e^{2x}} + 5\,\,gives \hfill \\
\frac{1}{2}\ln \,\left( {{e^{2x}} + 5} \right) + C \hfill \\
\end{gathered} \]
