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Math Help - Evaluate the Integral (Partial Fractions)

  1. #1
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    Evaluate the Integral (Partial Fractions)



    For the first problem, this is how far I get:
    http://www.rustybubble.net/mathwork.jpg
    (big image)

    But for the second, I don't know how to set it up.

    Thank you for your help!
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by birdz View Post


    For the first problem, this is how far I get:
    http://www.rustybubble.net/mathwork.jpg
    (big image)

    But for the second, I don't know how to set it up.

    Thank you for your help!
    For the second problem you need a fraction for each degree of (x-2)^2.

    So you need:

    \frac{A}{2x+1} + \frac{C}{x-2} + \frac{Dx+F}{(x-2)^2}<br />

    Good luck!
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  3. #3
    MHF Contributor Amer's Avatar
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    in the first integral you have two mistake the first one to decide A ,B ,C values cuz

    x^2(x-2)=a/x+b/x^2+c/(x-2)

    ax(x-2)+b(x-2)+cx^2=1
    Ax^2-2Ax+Bx-2B+Cx^2=1
    A+c=0
    B-2A=0
    -2B=1
    so
    B=-1/2 , A=-1/4 , C=1/4

    this is the first one the second
    the integrate of 1/x^2=-x^-1 not ln(x^2)

    thats all
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  4. #4
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    skeeter's Avatar
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    for repeated linear factors ...

    \frac{3x^2-9x+26}{(2x+1)(x-2)^2} = \frac{A}{2x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}<br />

    3x^2-9x+26 = A(x-2)^2 + B(2x+1)(x-2) + C(2x+1)

    let x = 2 ...

    20 = 5C

    C = 4

    let x = -\frac{1}{2} ...

    \frac{125}{4} = A \cdot \frac{25}{4}

    A = 5

    3x^2-9x+26 = 5(x-2)^2 + B(2x+1)(x-2) + 4(2x+1)

    let x = 0 ...

    26 = 20 - 2B + 4

    B = -1


    \frac{3x^2-9x+26}{(2x+1)(x-2)^2} = \frac{5}{2x+1} - \frac{1}{x-2} + \frac{4}{(x-2)^2}<br />
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