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?
$\displaystyle \frac{3\left(2x^2-8x-1\right)}{\left(x+4\right)\left(x+1\right)\left(2x-1\right)}=\frac{A}{\left(x+4\right)}+\frac{B}{\lef t(x+1\right)}+\frac{C}{\left(2x-1\right)}$
Multiply both sides by the common denominator to get $\displaystyle 3\left(2x^2-8x-1\right)=\left[A\left(x+1\right)\left(2x-1\right)\right]+\left[B\left(x+4\right)\left(2x-1\right)\right]+\left[C\left(x+4\right)\left(x+1\right)\right]$
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?
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.