Question

Taking constant of integration as zero, find f(1).

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

Answer 1

$I=\int \frac{x{e}^x}{{(x+1)}^2}dx=\int e^x\frac{(x+1-1)}{{(x+1)}^2}dx$
$=\int e^x[\frac{1}{x+1}+\frac{-1}{{(x+1)}^2}]dx$

Let $f\left(x\right)=\frac{1}{x+1}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{{(x+1)}^2}$

So, $I=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx.$
$=e^{x}f\left(x\right)+C$

$\Rightarrow I=e^x.\frac{1}{x+1}+C$

Given $C=0$

$\Rightarrow I=e^x.\frac{1}{x+1}=e^{x}f\left(x\right)$

And, $f\left(1\right)=\frac{1}{1+1}=\frac{1}{2}$