<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] $\int\frac{1}{x^2-a^2}\,\mathrm dx=\frac{1}{2a}\int\frac{1}{x-a}-\frac{1}{x+a}dx=\frac{1}{2a}(ln(x-a) -ln(x+a))$
and 
about $\int \frac{1}{x^2-6x}=\frac{-1}{6}\int \frac{1}{x}-\frac{1}{x-6}dx=\frac{-1}{6}(lnx-ln (x-6))$
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