Integrate (x^2+1)/(x^2-1)^2x2+1(x2−1)2 ?

(x^2+1)/(x^2-1)^2x2+1(x2−1)2

1 Answer
Ananda Dasgupta
Mar 22, 2018

-x/(x^2-1)+C−xx2−1+C

Explanation:

The denominator is of the form (x-1)^2(x+1)^2(x−1)2(x+1)2. Now,

(x-1)^2 +(x+1)^2 = 2(x^2+1)(x−1)2+(x+1)2=2(x2+1)

Thus

(x^2+1)/(x^2-1)^2 =1/2 {(x+1)^2+(x-1)^2}/{(x-1)^2(x+1)^2} = 1/2 1/(x-1)^2+1/2 1/(x+1)^2x2+1(x2−1)2=12(x+1)2+(x−1)2(x−1)2(x+1)2=121(x−1)2+121(x+1)2

So

int (x^2+1)/(x^2-1)^2 dx =1/2 int (1/(x-1)^2+ 1/(x+1)^2)dx ∫x2+1(x2−1)2dx=12∫(1(x−1)2+1(x+1)2)dx
qquad = 1/2(-1/(x-1)-1/(x+1))+C=-x/(x^2-1)+C

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