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To integrate (x^2+1)/(x^2-1)^2, use partial fractions and substitution.

First, factor the denominator: (x^2-1)^2 = (x-1)^2(x+1)^2

Then, use partial fractions to write the integrand as:

(x^2+1)/(x^2-1)^2 = A/(x-1) + B/(x-1)^2 + C/(x+1) + D/(x+1)^2

To find A, multiply both sides by (x-1) and let x=1:

A = lim(x->1) [(x^2+1)/(x+1)^2] = 1/4

Similarly, find B, C, and D:

B = lim(x->1) [(x^2+1)/(x+1)^3] = -1/8

C = lim(x->-1) [(x^2+1)/(x-1)^3] = 1/8

D = lim(x->-1) [(x^2+1)/(x-1)^2] = -1/4

Now, substitute the partial fractions into the integral:

∫(x^2+1)/(x^2-1)^2 dx = ∫[1/4(x-1) - 1/8(x-1)^2 + 1/8(x+1) - 1/4(x+1)^2] dx

Simplify and integrate each term:

= 1/4 ln|x-1| - 1/24(x-1)^3 + 1/8 ln|x+1| - 1/12(x+1)^3 + C

Therefore, the final answer is:

∫(x^2+1)/(x^2-1)^2 dx = 1/4 ln|x-1| - 1/24(x-1)^3 + 1/8 ln|x+1| - 1/12(x+1)^3 + C

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