Question
$\int e ^x \frac{x}{(x+1)^2} d x$

Answer

$\text { Let } I =\int e ^x\left(\frac{x}{(x+1)^2}\right) d x$
$=\int e ^x\left(\frac{(x+1)-1}{(x+1)^2}\right) d x$
$=\int e ^x\left(\frac{x+1}{(x+1)^2}-\frac{1}{(x+1)^2}\right) d x$
$=\int e ^x\left(\frac{1}{x+1}-\frac{1}{(x+1)^2}\right) d x$
$\text { Put } f ( x )=\frac{1}{x+1}$
$\therefore f ^{\prime}( x )=\frac{-1}{(x+1)^2}$
$\therefore \text { I }=\int e ^x\left[ f (x)+ f ^{\prime}(x)\right] d x$
$= e ^x \cdot f (x)+ c$
$\therefore I = e ^x\left(\frac{1}{x+1}\right)+ c $

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