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

1 Answer
Roy E.
Dec 19, 2016

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.

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