Integral Question - $\int\frac{1}{x^2-6x}\,\mathrm dx$ How I can evaluate the indefinite integral? :
 $$\int\frac{1}{x^2-6x}\,\mathrm dx$$ 
Do I need to bring it to this format? :   $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!

$$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$
$$1=A(x-6)+Bx$$
$$(A+B)x-6A=1$$
$$-6A=1,A+B=0 $$
$$A=-1/6,B=1/6$$
$$\int\frac{1}{x^2-6x}dx=\frac{1}{6}\int\left(\frac{1}{x-6}-\frac{1}{x}\right)dx=$$
$$=\frac{1}{6}(\ln|x-6|-\ln|x|)+C$$
