What's the integral of 1/(x^2+4)?

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

To find the integral of 1/(x^2+4), we can use the substitution method. Let u = x/2, then du/dx = 1/2 and dx = 2du. Substituting these into the integral, we get:

∫ 1/(x^2+4) dx = ∫ 1/((2u)^2+4) 2du
= (1/2) ∫ 1/(u^2+1) du

We can now use the formula for the integral of 1/(u^2+1), which is arctan(u) + C. Substituting back in for u, we get:

(1/2) ∫ 1/(u^2+1) du = (1/2) arctan(x/2) + C

Therefore, the integral of 1/(x^2+4) is (1/2)arctan(x/2) + C.