## Calculus: Early Transcendentals (2nd Edition)

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