Integrate #(x^2+1)/(x^2-1)^2# ?

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#(x^2+1)/(x^2-1)^2#

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Answer 1 · 2018-03-22T18:31:53

#-x/(x^2-1)+C#

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

#(x-1)^2 +(x+1)^2 = 2(x^2+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)^2#

So

#int (x^2+1)/(x^2-1)^2 dx =1/2 int (1/(x-1)^2+ 1/(x+1)^2)dx #
#qquad = 1/2(-1/(x-1)-1/(x+1))+C=-x/(x^2-1)+C#