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

Question

1479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-13T11:13:43

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

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

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

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

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

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

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

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