# intergration partial fraction

• Jan 31st 2009, 10:09 PM
walk_in_fark
intergration partial fraction
hi, i need help fast, i cant solve this

separate into partial fractions: 3(2x^2 - 8x -1)/(x +4)(x +1)(2x -1)
• Jan 31st 2009, 10:14 PM
Chris L T521
Quote:

Originally Posted by walk_in_fark
hi, i need help fast, i cant solve this

separate into partial fractions: 3(2x^2 - 8x -1)/(x +4)(x +1)(2x -1)

Are you familiar with the decomposition process?

$\frac{3\left(2x^2-8x-1\right)}{\left(x+4\right)\left(x+1\right)\left(2x-1\right)}=\frac{A}{\left(x+4\right)}+\frac{B}{\lef t(x+1\right)}+\frac{C}{\left(2x-1\right)}$

Multiply both sides by the common denominator to get $3\left(2x^2-8x-1\right)=\left[A\left(x+1\right)\left(2x-1\right)\right]+\left[B\left(x+4\right)\left(2x-1\right)\right]+\left[C\left(x+4\right)\left(x+1\right)\right]$

Expand the terms on the RHS of the equation. Then equate the corresponding coefficients.

Once you find the coefficients, then you can easily compute the integral.

Can you take it from here?
• Jan 31st 2009, 10:22 PM
Jhevon
alternatively, once you get to the form Chris has, plug in x = -4 to solve for A, then plug in x = -1 to solve for B and plug in x = 1/2 to solve for C
• Jan 31st 2009, 10:23 PM
walk_in_fark
yea, that is how i'm doing it, but when i derive an answer for A i keep getting -9 and the textbook says that the answer is; 7/(x + 4) - 3/(x +1) - 2/(2x -1). I've been trying to figure out where i've been going wrong for a while now to no avail.

It is pretty frustrating, maybe I should just go sleep lol(Headbang)

Seriously, thanks for all your help, I really appreciate it, I'm trying to teach myself math and it gets frustrating now and then.
• Jan 31st 2009, 10:46 PM
walk_in_fark
you ever have those moments where you just go D'oh. Well i just had one, i figured it out. It is really time for bed. Thanks for all your help guys, goodnight