# Partial Fractions Help

• Dec 14th 2013, 06:21 PM
wondering
Partial Fractions Help
$\int\frac{x-6}{x^2(x-2)}$

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

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

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

$3=B+C$

$0=A+B$

$1=C$

$A=-2$

$B=2$

$C=1$

I know this is wrong; where have I screwed up. Using the cover up method, I know that A should equal -1. Help. TIA.
• Dec 14th 2013, 06:41 PM
chiro
Re: Partial Fractions Help
Hey wondering.

I think your line should be x - 6 = Ax^2 + Bx(x-2) + C(x-2) which gives:

x - 6 = Ax^2 + Bx^2 - 2Bx + Cx - 2C which has a term 2Bx instead of 2B if you are using that particular decomposition listed above.
• Dec 14th 2013, 07:44 PM
wondering
Re: Partial Fractions Help
Thank you. It must be time to quit doing math for the night. :)