Answer
\[ = \frac{1}{2}\ln \,\left( {{e^{2x}} + 1} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^{2x}}}}{{{e^{2x}} + 1}}\,dx} \hfill \\
\hfill \\
u = {e^{2x}} + 1\,\,\,\,then\,\,\,du = 2{e^{2x}}dx \hfill \\
\hfill \\
apply\,\,the\,\,\,substitution \hfill \\
\hfill \\
= \frac{1}{2}\int_{}^{} {\frac{{du}}{u}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{2}\ln \left| u \right| + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = {e^{2x}} + 1 \hfill \\
\hfill \\
= \frac{1}{2}\ln \,\left( {{e^{2x}} + 1} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
