$\displaystyle \frac{6x^2 - 4x - 20}{x(x-4)(x+1)}dx$

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

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- Apr 20th 2010, 12:37 PMx5pyd3rxPartial Fraction.
$\displaystyle \frac{6x^2 - 4x - 20}{x(x-4)(x+1)}dx$

I have no clue how to proceed from here. everything i try dosent look right. - Apr 20th 2010, 12:43 PMharish21
Before starting integration, use partial fractions to decompose your term>

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

or,

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

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