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Math Help - Partial Fraction

  1. #1
    Junior Member
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    Partial Fraction

    integrate: \frac{x^3 + 3x^2 + 5x -9}{(x-5)(x+1)^3} dx


    I worked it down to x^3 + 3x^2 + 5x -9 = A(x+1)^3 + B(x-5)(x+1)^2 + c(x-5)(x+1) + D(x-5)

    and then

    x = 5
    a= 1

    x = -1
    -12 = b

    whats confusing me is that if i let x = -1 for c i would get -12 = C0 which is -12 = 0
    is my equation wrong?
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by x5pyd3rx View Post
    integrate: \frac{x^3 + 3x^2 + 5x -9}{(x-5)(x+1)^3} dx


    I worked it down to x^3 + 3x^2 + 5x -9 = A(x+1)^3 + B(x-5)(x+1)^2 + c(x-5)(x+1) + D(x-5)

    and then

    x = 5
    a= 1

    x = -1
    -12 = b

    whats confusing me is that if i let x = -1 for c i would get -12 = C0 which is -12 = 0
    is my equation wrong?
    Never thought about doing the question like that but lets roll with it...

    If x=5, everything bu the A term is 0 so we get...

    216A = 216.

    So A = 1.

    If x=-1, everything but the D term cancels and you get...

    -6D = -12, D = 2

    Now set x to be something generic like 0.

    Chances are you'll get something like ...B + ...C = -9

    Then set x to be something else. Like 1 for example.

    Once again you'll get an equation involving B and C.

    Solve for B and C using simultaneous equations.
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  3. #3
    Super Member Deadstar's Avatar
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    Alternatively you could see that you'll get Ax^3 + ... + Bx^3 + ...

    So B must be equal to 0.

    Something tells me that won't work though. You sure that equation is going to work..?
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  4. #4
    Super Member Deadstar's Avatar
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    Actually yeah it will be. C=0 as well.
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