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Math Help - help with partial fraction integral

  1. #1
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    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?
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  2. #2
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    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)
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  3. #3
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    would you mind showing me?
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  4. #4
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    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
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  5. #5
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    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.
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  6. #6
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    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
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  7. #7
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    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
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