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!
 A: $$\int\frac{1}{x^2-6x}\,\mathrm dx = \int\frac{1}{x(x-6)}\,\mathrm dx$$
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 
$$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx = \int \left(\color{blue}{\bf -}\frac{\color{blue}{\bf 1}}{\color{blue}{\bf 6}x} + \frac{\color{red}{\bf 1}}{\color{red}{\bf 6}(x-6)}\right)\,\mathrm dx$$
$$ = -\frac 16 \int \left(\dfrac 1x \right) \,\mathrm dx + \frac 16 \int \left(\frac 1{x-6}\right)\,\mathrm dx$$
A: The obvious method is partial fractions, but if you desperately want it in the form $\displaystyle \int \dfrac{1}{x^2-a^2}\, dx$ then complete the square on the denominator and substitute $u=x-3$; you get
$$\int \dfrac{1}{x^2-6x}\, dx = \int \dfrac{1}{u^2-9}\, du$$
A: 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
$$\int\frac{1}{x-a}dx=\ln|x-a|$$
A: $\int\frac{1}{x^2-a^2}\,\mathrm dx=\frac{1}{2a}\int\frac{1}{x-a}-\frac{1}{x+a}dx=\frac{1}{2a}(ln(x-a) -ln(x+a))$
and 
about $\int \frac{1}{x^2-6x}=\frac{-1}{6}\int \frac{1}{x}-\frac{1}{x-6}dx=\frac{-1}{6}(lnx-ln (x-6))$
A: $$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$
$$1=A(x-6)+Bx$$
$$(A+B)x-6A=1$$
$$-6A=1,A+B=0 $$
$$A=-1/6,B=1/6$$
$$\int\frac{1}{x^2-6x}dx=\frac{1}{6}\int\left(\frac{1}{x-6}-\frac{1}{x}\right)dx=$$
$$=\frac{1}{6}(\ln|x-6|-\ln|x|)+C$$
A: I guess I'm late)
Denote $x-3=t$
$$
\int \frac{dx}{x^2-6x}=\int \frac{dx}{x^2-6x +9-9}=\int \frac{dt}{(t-3)(t+3)}=\frac{1}{6} \int \frac{dt}{t-3} -\frac{1}{6} \int \frac{dt}{t+3}=\frac{1}{6}\log \Bigg|\frac{x-6}{x} \bigg|
$$
