What's the integral of x^3/(1+x^2)?

The integral of x^3/(1+x^2) is (1/2)ln(1+x^2) - (1/2)x^2 + C.

To solve this integral, we can use substitution. Let u = 1+x^2, then du/dx = 2x. Rearranging, we get dx = du/2x. Substituting these into the integral, we get:

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

Therefore, the integral of x^3/(1+x^2) is (1/2)ln(1+x^2) - (1/2)x^2 + C.