How do you integrate #int x^3/(x^2 -1)# using partial fractions?

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Answer 1 · 2016-06-24T18:46:37

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

I'm not sure you can do that

from some simple long division, turns out that # x^3/(x^2 -1) = x + x/(x^2 -1)#

so the integration is

#int \ x + x/(x^2 -1) \ dx# which is straightaway do-able

#int \ x + color{red}{x/(x^2 -1)} \ dx#
#= x^2/2 + 1/2 ln (x^2 -1) +C#

note the pattern in the red term of integrand, it is in form

#(\alpha f'(x))/(f(x))# and here #alpha = 1/2#, so you can "just do it"