Answer
\[ = - \frac{1}{{2\,\left( {1 + {e^{2x}}} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} \hfill \\
\hfill \\
rewrite\,\, \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} = \int_{}^{} {\frac{{{e^{2x}}}}{{\,{{\left( {1 + {e^{2x}}} \right)}^2}}}dx} \hfill \\
\hfill \\
Let\,,\,\,1 + {e^{2x}} = t\,\,\,\,\,\,then\,\,\,\,\,\,2{e^{2x}}dx = dt \hfill \\
\hfill \\
apply\,\,the\,\,substitution \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2}}}} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
= - \frac{1}{{2t}} + C \hfill \\
\hfill \\
substitute\,\,for\,\,\,\left( t \right) \hfill \\
\hfill \\
= - \frac{1}{{2\,\left( {1 + {e^{2x}}} \right)}} + C \hfill \\
\end{gathered} \]
