Show that

Let $\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}$

$1=A(x-3)+B(x+3)$

Equating the coefficients of x and constant term, we obtain

$A+B=0$

$-3 A+3 B=1$

On solving, we obtain

$A=-\frac{1}{6}$ and $B=\frac{1}{6}$

$\therefore \frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}$

$\Rightarrow \int \frac{1}{\left(x^{2}-9\right)} d x=\int\left(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\right) d x$

$=-\frac{1}{6} \log |x+3|+\frac{1}{6} \log |x-3|+\mathrm{C}$

$=\frac{1}{6} \log \left|\frac{(x-3)}{(x+3)}\right|+C$