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

The integral of 1/(x^2+1)^2 is (1/2)arctan(x/(x^2+1)) + C.

To solve this integral, we can use the substitution u = x^2 + 1. Then du/dx = 2x, so dx = du/(2x). Substituting these into the integral, we get:

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

Integrating u^(-2), we get -u^(-1) + C. Substituting back in for u, we get:

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

Simplifying, we get:

(1/2)arctan(x/(x^2+1)) + C

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