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Math Help - evaluate the integral using partial fractions

  1. #1
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    evaluate the integral using partial fractions

    The equation is integral sign ("S") (x^3 -4x^2 + 3x + 7)/(x^2 -3x +2) dx

    So first I did the long division and I got

    the original equation = S (x - 1 + (9-2x)/(x^2-3x+2)) dx

    so 9-2x = A + B
    ____ __ ___
    (x-1)(x-2) (x-1) (x-2)

    9-2x = A (x-2) + B(x-1)

    so if x = 1 then A = -7 and if x = 2 then B = 5

    which gives us

    S (x-1- 7/(x-2) + 5/(x-1)) dx = (1/2)x^2 -x -7 ln(x-2) + 5 ln (x-1) +C

    Did I do this correctly? I don't like ln but I am assuming I bring the costant infront?

    Thanks
    Calculus Beginner
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  2. #2
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    You are correct all the way until -7\ln|x-2|+5\ln|x-1|, which should be 5\ln|x-2|-7\ln|x-1|.

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  3. #3
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    Quote Originally Posted by calcbeg View Post
    The equation is integral sign ("S") (x^3 -4x^2 + 3x + 7)/(x^2 -3x +2) dx

    So first I did the long division and I got

    the original equation = S (x - 1 + (9-2x)/(x^2-3x+2)) dx

    so 9-2x = A + B
    ____ __ ___
    (x-1)(x-2) (x-1) (x-2)

    9-2x = A (x-2) + B(x-1)

    so if x = 1 then A = -7 and if x = 2 then B = 5

    which gives us

    S (x-1 + 5/(x-2) - 7/(x-1)) dx = (1/2)x^2 -x + 5ln|x-2| - 7ln|x-1| +C
    corrections
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