How do you integrate $\int \frac{x^{3}}{x^{2} - 1}$ $int x^3/(x^2 -1)$ using partial fractions?

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Answer 1 · 2016-06-24T18:46:37

$= \frac{x^{2}}{2} + \frac{1}{2} ln (x^{2} - 1) + C$ $= x^2/2 + 1/2 ln (x^2 -1) +C$

I'm not sure you can do that

from some simple long division, turns out that $\frac{x^{3}}{x^{2} - 1} = x + \frac{x}{x^{2} - 1}$ $x^3/(x^2 -1) = x + x/(x^2 -1)$

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"

How do you integrate int x^3/(x^2 -1)∫x3x2−1int x^3/(x^2 -1) using partial fractions?