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!

$$\int\frac{1}{x^2-6x}\,\mathrm dx = \int\frac{1}{x(x-6)}\,\mathrm dx$$
Now use partial fraction decomposition to obtain an integral of the form $$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx$$
Now all you need to do is to determine the (constant) values of $A$ and of $B$ so that $$ \left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{1}{x(x-6)}.$$
$$\left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{A(x-6) + B(x)}{x(x - 6)} \iff A(x - 6) + B(x) = 1\iff (A + B)x - 6A = 1 $$ 
$$\iff A+B = 0 \;\;\text{and}\;\; -6A = 1 \iff \color{blue}{\bf A = -\frac 16}, \;\;\text{and}\;\; B = - A \iff \color{red}{\bf B = \frac 16}$$
This gives us, then, Now use 
$$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx = \int \left(\color{blue}{\bf -}\frac{\color{blue}{\bf 1}}{\color{blue}{\bf 6}x} + \frac{\color{red}{\bf 1}}{\color{red}{\bf 6}(x-6)}\right)\,\mathrm dx$$
$$ = -\frac 16 \int \left(\dfrac 1x \right) \,\mathrm dx + \frac 16 \int \left(\frac 1{x-6}\right)\,\mathrm dx$$
