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

The integral of x/(1+x^2) is ln|1+x^2|/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/(1+x^2) dx = ∫(1/u) (du/2) = (1/2) ln|u| + C
= (1/2) ln|1+x^2| + C

Therefore, the integral of x/(1+x^2) is ln|1+x^2|/2 + C. We can check this answer by differentiating it using the chain rule:

d/dx [ln|1+x^2|/2] = 1/(1+x^2) * d/dx [1+x^2]
= 1/(1+x^2) * 2x
= x/(1+x^2)

This matches the original integrand, so our answer is correct.