$\displaystyle \int\frac{x-6}{x^2(x-2)}$

$\displaystyle \frac{x-6}{x^2(x-2)}=\frac{A}{x-2}+\frac{B}{x}+\frac{C}{x^2}$

$\displaystyle x-6=Ax^2+Bx^2-2B+Cx-2C$

$\displaystyle x-6=x^2(A+B)+Cx-2(B+C)$

$\displaystyle 3=B+C$

$\displaystyle 0=A+B$

$\displaystyle 1=C$

$\displaystyle A=-2$

$\displaystyle B=2$

$\displaystyle C=1$

I know this is wrong; where have I screwed up. Using the cover up method, I know that A should equal -1. Help. TIA.