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

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

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

Multiplying both sides by (x+1)(x-1), we get:

1 = A(x-1) + B(x+1)

Setting x=1, we get A=1/2. Setting x=-1, we get B=-1/2.

Therefore, we can rewrite 1/(x^2-1) as:

1/(x^2-1) = 1/2(x+1) - 1/2(x-1)

Integrating both sides, we get:

∫1/(x^2-1) dx = 1/2 ln|x+1| - 1/2 ln|x-1| + C

Simplifying, we get:

∫1/(x^2-1) dx = ln|(x+1)/(x-1)| + C

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