Question

How do you integrate `int x^3/[x(x^2+2x+1)]` using partial fractions?

Answer 1

`x-2ln|x+1|-1/(x+1)+C`

Explanation:

`x^3/[x(x^2+2x+1)] = x^2/(x+1)^2 =(1-1/(x+1))^2`
`qquad = 1-2/(x+1)+1/(x+1)^2`

Thus

`int x^3/[x(x^2+2x+1)]dx = int ( 1-2/(x+1)+1/(x+1)^2)dx`
`qquad = x-2ln|x+1|-1/(x+1)+C`