Question
Evalute : Find the primitive of $\frac{1}{1+e^x}$

Answer

Let I be the primitive of $\frac{1}{1+e^x}$
$
\text { Then } \begin{aligned}
& I=\int \frac{1}{1+e^x} d x \\
&=\int \frac{\left(\frac{1}{e^x}\right)}{\left(\frac{1+e^x}{e^x}\right)} d x=\int \frac{e^{-x}}{e^{-x}+1} d x \\
&=-\int \frac{-e^{-x}}{e^{-x}+1} d x \\
&=-\log \left|e^{-x}+1\right|+c \\
& \quad \ldots\left[\because \frac{d}{d x}\left(e^{-x}+1\right)=-e^{-x}\right. \text { and } \\
&\left.\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
\end{aligned}
$

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