Answer
\[ = \frac{{ - 1}}{{\,\left( {{e^x} + 1} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 2{e^x} + 1}}\,dx} \hfill \\
\hfill \\
factoring \hfill \\
\hfill \\
= \int_{}^{} {\frac{{{e^x}}}{{\,{{\left( {{e^x} + 1} \right)}^2}}}\,dx} \hfill \\
\hfill \\
set\,\,{e^x} + 1 = t\,\,\,\, \to \,\,\,\,{e^x}dx = dt \hfill \\
\hfill \\
integral\,becomes \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 2{e^x} + 1}}\,dx} = \int_{}^{} {\frac{{dt}}{{{t^2}}}} \hfill \\
\hfill \\
\int_{}^{} {{t^{ - 2}}dt} \hfill \\
\hfill \\
{\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
- \frac{1}{t} + C \hfill \\
\hfill \\
substituting\,back\,\,t = {e^x} + 1 \hfill \\
\hfill \\
= \frac{{ - 1}}{{\,\left( {{e^x} + 1} \right)}} + C \hfill \\
\end{gathered} \]
