# Thread: help with partial fraction integral

1. ## help with partial fraction integral

I need some help finding C & D in the following integral
$\int\frac{2x^3-4x-8}{(x^2-x)(x^2+4)}$

$2x^3-4x-8=A(x-1)(x^2+4)+B(x)(x^2+4)+[C(2x)+D](x)(x-1)$

I used x=1 to find that B=-2
I used x=0 to find that A=2

I am now trying to find C & D and am having some trouble.

When I attempt to use x=-1, I get
$-6=-10+-4C+2D$

How do I find C & D? Is x=-1 a poor choice? What would be a better choice?

2. Hi

You have basically 2 ways to find C and D :
- expand and identify the coefficients of the powers of x
- choose 2 different values for x as you have done to find A and B (x=-1 is not a bad choice but you need another value to get a second equation)

3. would you mind showing me?

4. OK

1st method : expand

$2x^3-4x-8=2(x-1)(x^2+4)-2x(x^2+4)+(Cx+D)x(x-1)$

$2x^3-4x-8=(2x^3-2x^2+8x-8)-(2x^3+8x)+(Cx^3+Dx^2-Cx^2-Dx)$

$2x^3-4x-8=Cx^3+(D-C-2)x^2-Dx-8$

Therefore
2 = C
0 = D-C-2
-4 = -D
-9 = -8

2nd method : choose 2 values

$2x^3-4x-8=2(x-1)(x^2+4)-2x(x^2+4)+(Cx+D)x(x-1)$

For instance x=-1
-6 = -10-2C+2D

And x=2
0 = -16+4C+2D
which is the second equation

Note that I used Cx+D and not C(2x)+D as I understood you did

5. sorry to be a pain but I am still having some trouble, (my algebra is rusty)

I now have

-6 = -10 + -4C + 2D

-24 = 8C + 4D

Now what do I do.

6. With the first method you get directly C = 2 and D = 4

With the second method you get 2 linear equations
-6 = -10-2C+2D
0 = -16+4C+2D

You can solve using linear combination
For instance :
2nd equation - 1st equation gives : 6 = -6 + 6C therefore C = 2
Substituting C = 2 into 1st or 2nd equation gives D = 4

7. So we get
$\frac{2x^3-4x-8}{(x^2-x)(x^2+4)} = \frac{2}{x} - \frac{2}{x-1} + \frac{2x+4}{x^2+4}$

If you have a calculator the best way to be sure of your result is to calculate both sides using $x = \pi$, since $\pi$ is a transcendental number