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

Dec 19, 2016

$\frac{1}{6} \ln | \frac{x - 3}{x + 3} | + C$

#### Explanation:

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

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 $\frac{1}{x - a}$, substitute $u = x - 3$ in the first and $u = x + 3$ in the second.