<s>[INST] 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!
 [/INST] Use partial fraction decomposition to obtain a sum of easily integrable terms:
$$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$
$$\text{multiply by }x \implies \frac{1}{x-6}=A+\frac{Bx}{x-6},\quad \text{set }x=0 \implies \color{blue}{A=-\frac{1}{6}}$$
$$\text{multiply by }x-6 \implies \frac{1}{x}=\frac{A(x-6)}{x}+B,\quad \text{set }x=6 \implies \color{red}{B=\frac{1}{6}}$$
Hence
$$\int\frac{1}{x^2-6x}dx=\int\left(\color{blue}-\frac{\color{blue}1}{\color{blue}6x}+\frac{\color{red}1}{\color{red}6(x-6)}\right)dx$$
Now, use the fact that
$$\int\frac{1}{x-a}dx=\ln|x-a|$$
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