Question

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

Answer 1

`1/6ln|(x-3)/(x+3)|+C`

Explanation:

First cancel out the silly common factor `x^2-1`:
`int 1/((x-3)(x+3))dx`
Then split the denominator, leaving gaps to be filled in:
`int { } /(x-3)+{}/(x+3)dx`
Then use the cover-up rule of partial fractions to fill in the gaps:
`int { 1/(3+3)} /(x-3)+{1/(-3-3)}/(x+3)dx`
Tidy up:
`1/6 int1/(x-3)-1/(x+3)dx`
and do the standard integrals:
`1/6 (ln|x-3|-ln|x+3|)+C`
and optionally re-arrange use the properties of logarithms `log a - log b = log (a/b)` for any base:
`1/6 ln|(x-3)/(x+3)|+C`

In the cover up rule, you put your finger over the `x-3` in the denominator, and replace `x` with whatever value makes the expression under your finger equal to zero and evaluate the resulting small sum. Clearly this is `3`. Do the same for the other
denominator. If you don't accept the standard integral of `1/(x-a)`, substitute `u=x-3` in the first and `u=x+3` in the second.