Answer
\[ = - \frac{1}{2}\ln \left( {2 + {e^t}} \right) + \frac{1}{2}t + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dt}}{{2 + {e^t}}}} \hfill \\
\hfill \\
Given\,\,funcion \hfill \\
\hfill \\
\frac{1}{{2 + {e^t}}} \hfill \\
\hfill \\
rewriting \hfill \\
\hfill \\
\frac{1}{{2 + {e^t}}} = \frac{{ - \frac{1}{2}{e^t} + \frac{1}{2}{e^t} + 1}}{{2 + {e^t}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= - \frac{1}{2}\frac{{{e^t}}}{{2 + {e^t}}} + \frac{{\frac{1}{2}{e^t} + 1}}{{2 + {e^t}}} \hfill \\
\hfill \\
= - \frac{1}{2}\frac{{{e^t}}}{{2 + {e^t}}} + \frac{1}{2} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{dt}}{{2 + {e^t}}}} = \int_{}^{} {\,\left( { - \frac{1}{2}\frac{{{e^t}}}{{2 + {e^t}}} + \frac{1}{2}} \right)} \,dt \hfill \\
\hfill \\
write\,\,as\,\,two\,\,{\text{integrals}} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( { - \frac{1}{2}\frac{{{e^t}}}{{2 + {e^t}}}} \right)} dt + \int_{}^{} {\,\left( {\frac{1}{2}} \right)} dt \hfill \\
\hfill \\
let\,\,,\,\,u = 2 + {e^t}\,\,\,,\,\,then \hfill \\
\hfill \\
= - \frac{1}{2}\ln \left| u \right| + \frac{1}{2}t + C \hfill \\
\hfill \\
subsitute\,\,back\,\,u = 2 + {e^t} \hfill \\
\hfill \\
= - \frac{1}{2}\ln \left( {2 + {e^t}} \right) + \frac{1}{2}t + C \hfill \\
\hfill \\
\end{gathered} \]
