# Partial Fractional Decomposition to obtain an integral.

1. Jan 28, 2010

### Perseverance

1. The problem statement, all variables and given/known data
Find the exact values of A, B, and C in the following partial fraction decomposition.Then obtain the integral using those values.
1/(x^3-3x^2) = A/x + B/x^2 + C/(x-3)

2. Relevant equations
I'm not sure at this point.

3. The attempt at a solution
To make the denominator on the right equal to the left...
A(x^2)(x-3) + B(x)(x-3) + C(x)(x^2)
so.....
1 = Ax^3 +Cx^3 -3Ax^2 +Bx^2 -3Bx
From cubic values..
A + C = 0 So A= -C
From squared values...
-3A + B = 0 So A = 1/3B which also means C = -1/3B
From x values..
-3B = 0.

Here is where im at a loss. The x value equation cannot possibly be correct. It would denote that B=0, therefore A and C also equal 0. Am I doing something incorrectly in my multiplication? If not, how is it possible to use the partial fractional decomposition method to make an equation for integration?

2. Jan 28, 2010

### zcd

you made the mistake here: $$1=A(x)(x-3)+B(x-3)+C(x^{2})$$

3. Jan 28, 2010

### Perseverance

Wow, this problem just became possible. Thank you so much :)

4. Jan 28, 2010

### Perseverance

Actually, what implications does factoring out an x have on the numerical answers?

Edit: Never mind I understand how that was able to be done, it should have no implications. thanks again!

5. Jan 28, 2010

### HallsofIvy

Staff Emeritus
Multiplying both sides of your first equation by x^2(x-3) gives
1= Ax(x-3)+ B(x-3)+ Cx^2
Taking x= 0 gives immediately 1= -3B and taking x= 3 gives immediately 1= 9C. Taking x= 1 (just because it is a simple number) gives 1= -2A- 2B+ C. Since you already know B and C, that is easy to solve for A.