The question is to evaluate the integral x/x-6 dx by partial fractions

This is how I've started

$\displaystyle

\begin{array}{l}

\int {\frac{x}{{x - 6}}dx} \\

= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\

\end{array}

$

so giving

$\displaystyle

\begin{array}{l}

= \int {\frac{{x - 1 + 1}}{{x - 6}}dx = \int {\frac{{x - 1}}{{x - 6}} + \frac{1}{{x - 6}}} } \\

\frac{A}{{x - 6}} + \frac{B}{{x - 6}} = \frac{{x - 1 + 1}}{{x - 6}} \\

A(x - 6)(x - 6) + B(x - 6)(x - 6) = x(x - 6) \\

A(x^2 - 12x + 36) + B(x^2 - 12 + 36) = x^2 - 6x \\

\end{array}

$

rearranging this to

$\displaystyle

(A + B)x^2 + ( - 12A - 12B)x + (A + B)36 = x^2 - 6x

$

and then I have

A+B = 1 due to the co-efficient of x^2 term of polynomial both sides

-12A-12B= -6 due to the co-efficient of x

and A+B = 0 due to the constant

I'm really having trouble on how to solve for these .