Asked by abc

¡ìdx/(x^2-1)^2
Answered by Damon
(1/4)[-2x/(x^2-1) - ln(1-x)+ln(1+x) ] + c

used Wolfram alpha
Answered by Steve
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)
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