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

The integral of x/(x^2+1) is ln|x^2+1|/2 + 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:

∫x/(x^2+1) dx = ∫(1/2) * (x/(x^2+1)) * 2x dx
= ∫(1/2) * (1/u) du
= (1/2) ln|u| + C
= (1/2) ln|x^2+1| + C

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

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