How to integrate x^4/(x^2+1)^2?

To integrate x^4/(x^2+1)^2, use substitution with u = x^2+1.

To integrate x^4/(x^2+1)^2, use substitution with u = x^2+1. Then, du/dx = 2x and dx = du/2x. Substituting these into the integral gives:

∫x^4/(x^2+1)^2 dx = ∫(u-1)^2/u^2 du/2x
= 1/2 ∫(u^(-1) - 2u^(-2) + u^(-3)) du

Integrating each term separately gives:

= 1/2 ln|u| + u^(-1) - u^(-2)/2 + C
= 1/2 ln|x^2+1| + (x^2+1)^(-1) - (x^2+1)^(-2)/2 + C

Therefore, the final answer is:

∫x^4/(x^2+1)^2 dx = 1/2 ln|x^2+1| + (x^2+1)^(-1) - (x^2+1)^(-2)/2 + C