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