How to integrate ((x^2)-1)^-1.....help me plizzz....

Note that:
(x^2 - 1)^(-1) = 1/(x^2 - 1).

Since x^2 - 1 factors to (x + 1)(x - 1), we can use Partial Fractions to write:
1/(x^2 - 1) = A/(x + 1) + B/(x - 1), for some A and B.

Multiplying both sides by (x + 1)(x - 1) gives:
1 = A(x - 1) + B(x + 1).

(i) Letting x = -1: -2A = 1 ==> A = -1/2
(ii) Letting x = 1: 2B = 1 ==> B = 1/2.

Thus:
1/(x^2 - 1) = (-1/2)/(x + 1) + (1/2)/(x - 1)
==> 1/(x^2 - 1) = 1/[2(x - 1)] - 1/[2(x + 1)].

Integrating term-by-term yields:
∫ 1/(x^2 - 1) dx
= ∫ {1/[2(x - 1)] - 1/[2(x + 1)]} dx
= 1/2 ∫ 1/(x - 1) dx - 1/2 ∫ 1/(x + 1) dx
= (1/2)ln|x - 1| - (1/2)ln|x + 1| + C.

I hope this helps!

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

Integrating we have
(1/2){ln(|x - 1|) + c1 - ln(|x + 1|) - c2}, by partial fractions
= (1/2){ln(|x - 1|) - ln(|x + 1|) +C}

((x^2)-1)^-1 = 1/(x^2 - 1) = (1/2)[1/(x-1) - 1/(x+1)]
So, the integral = (1/2)(ln|x-1| - ln|x+1|) + c