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

Answer

$ \text { Let } I =\int \frac{1+x}{x+ e ^{-x}} d x$
$=\int \frac{1+x}{x+\frac{1}{ e ^x}} d x$
$=\int \frac{1+x}{\frac{x \cdot e ^x+1}{ e ^x}} d x$
$=\int \frac{ e ^x(1+x)}{x \cdot e ^x+1} d x $
Put $x \cdot e^x+1=t$
$ \therefore\left[x \cdot\left(e^x\right)+e^x \cdot(1)+0\right] d x=d t$
$\therefore e^x(x+1) d x=d t$
$\therefore I=\int \frac{d t}{t}$
$=\log |t|+c $
$\therefore I =\log \left| x \cdot e ^{ x }+1\right|+ c$

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