¡ìdx/(x^2-1)^2

Answer 1

(1/4)[-2x/(x^2-1) - ln(1-x)+ln(1+x) ] + c

used Wolfram alpha

Answer 2

And you arrive there by using partial fractions:

1/(x^2-1)^2 = 1/4 (1/(x+1) + 1/(x+1)^2 - 1/(x-1) + 1/(x-1)^2)