Question

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

Answer 1

`int [(2x^3-x^2)*dx]/(x^2+1)^2`

=`int (2x^2-x)/4*(4x*dx)/(x^2+1)^2`

=`(2x^2-x)/4*-1/(x^2+1)`-`int [(4x-1)/4*dx]*(-1)/(x^2+1)`

=`-(2x^2-x)/(4x^2+4)`+`1/4`*`int [(4x-1)*dx]/(x^2+1)`

=`-(2x^2-x)/(4x^2+4)`+`1/2``int (2x*dx)/(x^2+1)`-`1/4``int dx/(x^2+1)`

=`-(2x^2-x)/(4x^2+4)+1/2ln(x^2+1)-1/4*arctanx+C`

Explanation:

I used integration by parts with `u=(2x^2-x)/4` and `dv=(4x*dx)/(x^2+1)^2`