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

Answer

$\text { Let } I =\int e ^x \frac{\left(1+x^2\right)}{(1+x)^2} d x$
$=\int e ^x\left[\frac{x^2-1+2}{(1+x)^2}\right] d x$
$=\int e ^x\left[\frac{x^2-1}{(x+1)^2}+\frac{2}{(x+1)^2}\right] d x$
$=\int e ^x\left[\frac{x-}{x+1}+\frac{2}{(x+1)^2}\right] d x$
Put $f ( x )=\frac{x-1}{x+1}$
$\therefore f ^{\prime}( x )=\frac{(x+1)(1-0)-(x-1)(1+0)}{(x+1)^2}$
$=\frac{2}{(x+1)^2}$
$\therefore I =\int e ^x\left[ f (x)+ f ^{\prime}(x)\right] d x$
$= e ^{ x } \cdot f ( x )+ c$
$= e ^x\left(\frac{x-1}{x+1}\right)+ c$

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