Let $I=\int \frac{dx}{x^2-9}$Let $\frac{1}{x^2-9}=\frac{1}{(x-3)(x+3)}$ $=\frac{A}{x-3}+\frac{B}{x+3}$$\Rightarrow 1 \equiv A(x+3)+B(x-3)$...(i)Putting $x=3$ in (i), we get $1=A(3+3)$ $\Rightarrow A=\frac{1}{6}$Putting $x=-3$ in (i), we get $1=B(-3-3)$ $\Rightarrow B=-\frac{1}{6}$$\therefore \int \frac{dx}{x^2-9}$ $=\frac{1}{6}\int \left[\frac{1}{x-3}-\frac{1}{x+3}\right]dx$ $=\frac{1}{6}\left[\log|x-3|-\log|x+3|+C\right]$ $=\frac{1}{6}\log\left|\frac{x-3}{x+3}\right|+C$