What's the integral of x/(x^2+4)?

The integral of x/(x^2+4) is 1/2 ln(x^2+4) + C.

To solve this integral, we can use the substitution method. Let u = x^2 + 4, then du/dx = 2x. Rearranging, we get dx = du/2x. Substituting these into the integral, we get:

∫ x/(x^2+4) dx = ∫ (1/2) * (2x/(x^2+4)) dx
= (1/2) ∫ du/u
= (1/2) ln|u| + C
= (1/2) ln|x^2+4| + C

Therefore, the integral of x/(x^2+4) is 1/2 ln(x^2+4) + C. It is important to note that the absolute value is used in the natural logarithm because the argument (x^2+4) can be negative for some values of x.