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Partial Fractional Decomposition to obtain an integral.

  1. Jan 28, 2010 #1
    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)
    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. jcsd
  3. Jan 28, 2010 #2


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    you made the mistake here: [tex]1=A(x)(x-3)+B(x-3)+C(x^{2})[/tex]
  4. Jan 28, 2010 #3
    Wow, this problem just became possible. Thank you so much :)
  5. Jan 28, 2010 #4
    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!
  6. Jan 28, 2010 #5


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    Science Advisor

    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.

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