# Fraction integral need help!

• May 17th 2009, 07:23 PM
DHS1
Fraction integral need help!
$\int\frac{3x}{(x-1)(x^2+6)}$

I got this far-

$\int\frac{3x}{(x-1)(x^2+6)} = \frac{A}{x-1}+\frac{Bx+C}{x^2+6}$

But after that I don't know how to solve for it because of the Bx+C

Any help is appreciated! Test tomorrow and I need to know this...
• May 17th 2009, 08:16 PM
TwistedOne151
Start from
$\frac{3x}{(x-1)(x^2+6)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+6}$
multiply by the common denominator $(x-1)(x^2+6)$:
$3x=A(x^2+6)+(Bx+C)(x-1)$
$3x=Ax^2+6A+Bx^2-Bx+Cx-C$
$3x=(A+B)x^2+(C-B)x+(6A-C)$
Now equate coefficients.

--Kevin C.