What is the integral of 1/ (x^2-4)?

Answer 1

This is a good application of partial fractions:

1/(x²-4) = 1/(x+2)(x-2) = A/(x+2) + B/(x-2) =

A(x-2) + B(x+2) = 1

Ax - 2A + Bx + 2B = 1

(A+B)x + 2(B-A) = 1

A+B = 0

2(B-A) = 1

Solving the system of equations, you get A = -1/4 and B = 1/4. So:

∫(1/(x²-4) dx = -(1/4)∫1/(x+2) dx + (1/4)∫1/(x-2) dx

You can solve the two integrals on the RHS above using simple substitution: u = x+2 and v = x-2.