Partial Fraction.

• Apr 20th 2010, 01:37 PM
x5pyd3rx
Partial Fraction.
$\frac{6x^2 - 4x - 20}{x(x-4)(x+1)}dx$

I have no clue how to proceed from here. everything i try dosent look right.
• Apr 20th 2010, 01:43 PM
harish21
Quote:

Originally Posted by x5pyd3rx
$\frac{6x^2 - 4x - 20}{x(x-4)(x+1)}dx$

I have no clue how to proceed from here. everything i try dosent look right.

Before starting integration, use partial fractions to decompose your term>

$\frac{6x^2 - 4x - 20}{x(x-4)(x+1)} = \frac{A}{x} + \frac{B}{x-4} + \frac{C}{x+1}$.....................(1)

or,

${6x^2 - 4x - 20}= A(x-4)(x+1) + B (x)(x+1) + C (x)(x-4)$

find out A, B, and C. Plug them in equation (1) and then integrate!