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

Answer

$
\begin{aligned}
LetI & =\int e^x \cdot \frac{x}{(x+1)^2} d x \\
& =\int e^x\left[\frac{(x+1)-1}{(x+1)^2}\right] d x \\
& =\int e^x\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right] d x
\end{aligned}
$
Put $f(x)=\frac{1}{x+1}$
Then $f^{\prime}(x)=\frac{d}{d x}(x+1)^{-1}=-1(x+1)^{-2} \cdot \frac{d}{d x}(x+1)$
$
\begin{aligned}
& =\frac{-1}{(x+1)^2} \times(1+0)=\frac{-1}{(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 \cdot \frac{1}{x+1}+c .
\end{aligned}
$

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