Question

How do you integrate `x/(x+1)^3`?

Answer 1

The answer is `=-1/(x+1)+1/(2(x+1)^2)+C`

Explanation:

Perform this integral by substitution

Let `u=x+1`, `=>0`, `du=dx`

The integral is

`I=int(xdx)/(x+1)^3=int((u-1)du)/u^3`

`=int(u^-2-u^-3)du`

`=-1/u+1/(2u^2)`

`=-1/(x+1)+1/(2(x+1)^2)+C`

Answer 2

`I=-(2x+1)/(2(x+1)^2)+c`

Explanation:

Here,

`I=intx/(x+1)^3dx`

Subst . `color(blue)(x+1=u=>x=u-1=>dx=du`

So,

`I=int(u-1)/u^3du`

`=int[1/u^2-1/u^3]du`

`=int[u^-2 -u^-3 ]du`

`=u^(-2+1)/(-2+1)-u^(-3+1)/(-3+1)+c`

`=u^-1/(-1)-u^-2/(-2)+c`

`=-1/u+1/(2u^2)+c`

`=1/(2u^2)-1/u+c`

`=(1-2u)/(2u^2)+c`

Subst. back `color(blue)(u=x+1`

`I=(1-2x-2)/(2(x+1)^2)+c`

`I=-(2x+1)/(2(x+1)^2)+c`