Question
Find $\int \frac { \left( x ^ { 2 } + 1 \right) e ^ { x } } { ( x + 1 ) ^ { 2 } } d x.$
$= e ^ { x } - 2 \int e ^ { x } \left( \frac { x+1 } { ( x + 1 )^2 } + \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } } \right) d x$
$= e ^ { x } - 2 \int e ^ { x } \left( \frac { 1 } { ( x + 1 ) } + \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } } \right) d x$
Consider $f ( x ) = \frac { 1 } { x + 1 },$$\therefore$$f ^ { \prime } ( x ) = \frac { ( - 1 ) } { ( x + 1 ) ^ { 2 } }$
Thus, the above integral is of the form
$\because \int e ^ { x } f ( x ) + f ^ { \prime } ( x ) d x = e ^ { x } f ( x ) + C $
$\therefore \quad I = e ^ { x } - 2 e ^ { x }\Big[ \frac { 1 } { ( x + 1 ) } \Big]+ C$
$\Rightarrow I = e ^ { x } \left( \frac { x + 1 - 2 } { x + 1 } \right) + C $
$\Rightarrow I = e ^ { x } \left( \frac { x - 1 } { x + 1 } \right) + C$
$\therefore \int e ^ { x } \frac { \left( x ^ { 2 } + 1 \right) } { ( x + 1 ) ^ { 2 } } d x = e ^ { x } \left( \frac { x - 1 } { x + 1 } \right) + C$
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