# Thread: intergration partial fraction

1. ## 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)

2. 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?

3. 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

4. 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

Seriously, thanks for all your help, I really appreciate it, I'm trying to teach myself math and it gets frustrating now and then.

5. 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