Answer
\[ = {\tan ^{ - 1}}\,\left( 2 \right) - \frac{\pi }{4}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\ln 2} {\frac{{{e^x}}}{{1 + {e^{2x}}}}} \hfill \\
\hfill \\
so\,,\,\int_0^{\ln 2} {\frac{{{e^x}}}{{1 + \,{{\left( {{e^x}} \right)}^2}}}} \,\,dx \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \,\,\left[ {{{\tan }^{ - 1}}\,{e^x}} \right]_0^{\ln 2} \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
\,\,\left[ {{{\tan }^{ - 1}}{e^{\ln 2}}} \right] - \,\,\left[ {{{\tan }^{. - }}{e^0}} \right] \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
= {\tan ^{ - 1}}\,\left( 2 \right) - {\tan ^{ - 1}}\,\left( 1 \right) \hfill \\
\hfill \\
= {\tan ^{ - 1}}\,\left( 2 \right) - \frac{\pi }{4} \hfill \\
\end{gathered} \]
