Question

Integrate the function:
$\frac{x{e}^x}{(1+x{)}^2}$

Answer 1

To find: $\int \frac{x{e}^x}{(1+x{)}^2}.dx$

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

$=\int e^x\left(\frac{x+1-1}{(1+x{)}^2}\right).dx$

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

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

Using $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)=e^{x}f\left(x\right)+C$

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

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

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

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

Where $C$ is constant of integration.