Answer
\[ = \ln \,\left( {{e^x} + 1} \right) + C\]
Work Step by Step
\[\begin{gathered}
= \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}dx} \hfill \\
\hfill \\
set\,\,{e^x} + 1 = u\,\,then\,\,\,{e^x}dx = du \hfill \\
\hfill \\
integral\,\,becomes \hfill \\
\hfill \\
= \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}dx} = \int_{}^{} {\frac{{du}}{u}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \ln u + C \hfill \\
\hfill \\
substituting\,back\,\,\,u = {e^x} + 1 \hfill \\
\hfill \\
= \ln \,\left( {{e^x} + 1} \right) + C \hfill \\
\end{gathered} \]
