Question
Evalute : $\int \frac{1+x}{x+e^{-x}} d x$

Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{1+x}{x+e^{-x}} d x \\
& =\int \frac{(1+x) e^x}{\left(x+e^{-x}\right) e^x} d x \\
& =\int \frac{(1+x) e^x}{x e^x+1} d x
\end{aligned}
$
Put $x e^x+1=t$
$
\begin{aligned}
& \therefore\left(x e^x+e^x \times 1\right) d x=d t \\
& \therefore(1+x) e^x d x=d t \\
& \therefore I=\int \frac{1}{t} d t=\log |t|+c \\
& \quad=\log \left|x e^x+1\right|+c .
\end{aligned}
$

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