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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 and 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. It is important to note that the absolute value of x^2-1 is taken in the natural logarithm to account for negative values of x.

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