How do you find the integral of (1 over 1-x^2)

I know it must be pretty basic, but iv just gone blank for the moment

Trig substitution perhaps?

Yes, use trig substitution:

x = sinu
dx = cosu du

∫ 1 / (1 − x²) dx
= ∫ 1 / (1 − sin²u) * cosu du
= ∫ 1/cos²u * cosu du
= ∫ 1/cosu du
= ∫ secu du
= ln|tanu + secu| + C

Now we substitute back:
sinu = x
cosu = √(1−sin²u) = √(1−x²)
tanu = x/√(1−x²)
secu = 1/√(1−x²)

∫ 1 / (1 − x²) dx
= ln |x/√(1−x²) + 1/√(1−x²) | + C
= ln |x + 1| − ln (√(1−x²)) + C
= ln |x + 1| − 1/2 ln |1−x²| + C