Question

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

Answer 1

The correct answer is: $∫\frac{xe^x}{(1+x)^2}dx=\frac{e^x}{1+x}+C$
Let $I=∫\frac{xe^x}{(1+x)^2} dx=∫e^x{\frac{x}{(1+x)^2}}dx$
$=∫e^x=[\frac{1+x-1}{(1+x)^2}]dx$
$=∫e^x=[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx$
Let $ƒ(x)=\frac{1}{1+x}\,\, ƒ'(x)=\frac{-1}{(1+x)^2}$
$⇒∫\frac{xe^x}{(1+x)^2}dx=∫e^x[{ƒ(x)+ƒ'(x)}]dx$
It is known that,$∫e^x[ƒ(x)+ƒ'(x)]dx=e^xƒ(x)+C$
$∴∫\frac{xe^x}{(1+x)^2}dx=\frac{e^x}{1+x}+C$