What's the integral of 1/(x^2-9)?

The integral of 1/(x^2-9) is (1/6)ln|(x-3)/(x+3)| + C.

To solve this integral, we can use partial fractions. First, we factor the denominator as (x-3)(x+3). Then, we write 1/(x^2-9) as A/(x-3) + B/(x+3), where A and B are constants.

Multiplying both sides by (x-3)(x+3), we get 1 = A(x+3) + B(x-3). We can solve for A and B by setting x=3 and x=-3, respectively.

When x=3, we get 1 = 6A, so A=1/6. When x=-3, we get 1 = -6B, so B=-1/6.

Now we can rewrite the integral as ∫(1/6)(1/(x-3)) - (1/6)(1/(x+3)) dx.

Integrating each term separately, we get (1/6)ln|(x-3)/(x+3)| + C, where C is the constant of integration.

Therefore, the integral of 1/(x^2-9) is (1/6)ln|(x-3)/(x+3)| + C.