Integrate the rational function: $\frac{1}{x^{2}-9}$

Let $\frac{1}{x^{2}-9} = \frac{1}{(x+3)(x-3)}$.
Using partial fractions,we write $\frac{1}{(x+3)(x-3)} = \frac{A}{x+3} + \frac{B}{x-3}$.
Multiplying both sides by $(x+3)(x-3)$,we get $1 = A(x-3) + B(x+3)$.
Setting $x = 3$,we get $1 = B(6)$,so $B = \frac{1}{6}$.
Setting $x = -3$,we get $1 = A(-6)$,so $A = -\frac{1}{6}$.
Thus,$\int \frac{1}{x^{2}-9} dx = \int \left( \frac{-1/6}{x+3} + \frac{1/6}{x-3} \right) dx$.
$= -\frac{1}{6} \int \frac{1}{x+3} dx + \frac{1}{6} \int \frac{1}{x-3} dx$.
$= -\frac{1}{6} \ln |x+3| + \frac{1}{6} \ln |x-3| + C$.
$= \frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C$,where $C$ is the constant of integration.