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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    I got an answer but I'm not sure if i did the problem correctly.

    Evaluate the following integral using partial fractions:

    The integral of (x+1)/(x^2 - 5x + 6)

    My answer was -3ln(x-2)+(4/3)ln(x-3) + C

    Is this correct???
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by abcocoa View Post
    I got an answer but I'm not sure if i did the problem correctly.

    Evaluate the following integral using partial fractions:

    The integral of (x+1)/(x^2 - 5x + 6)

    My answer was -3ln(x-2)+(4/3)ln(x-3) + C

    Is this correct???
    Almost! the coefficient of ln(x - 3) is just 4, not 4/3

    (x+1)/(x^2 - 5x + 6) = (x+1)/(x - 3)(x - 2)

    so we have (x+1)/(x - 3)(x - 2) = A/(x - 3) + B/(x - 2)
    multiply through by (x - 3)(x - 2) we obtain

    x + 1 = A(x - 2) + B(x - 3)
    => x + 1 = Ax - 2A + Bx - 3B
    => x + 1 = (A + B)x + (-2A - 3B)
    => A + B = 1..........................(1)
    and -2A - 3B = 1 ....................(2)

    2A + 2B = 2 ................(3) = (1)*2
    -2A - 3B = 1 ...............(2)

    => -B = 3
    => B = -3

    => A - 3 = 1
    => A = 4

    so (x+1)/(x^2 - 5x + 6) = 4/(x - 3) - 3/(x -2)
    so int{(x+1)/(x^2 - 5x + 6)}dx = int{4/(x - 3) - 3/(x -2)}dx
    = 4ln(x - 3) - 3ln(x - 2) + C ...........so you were almost correct.

    Remember, whenever you integrate and get an answer, you can check if you're right by differentiating the answer you got.

    d/dx(4ln(x - 3) - 3ln(x - 2) + C) = 4*(1/x - 3) - 3*(1/(x - 2))
    = 4/(x - 3) - 3/(x - 2) + 0
    = [4(x - 2) - 3(x - 3)]/[(x - 3)(x - 2)]
    = (4x - 8 - 3x + 9)/(x - 3)(x - 2)
    = (x + 1)/(x - 3)(x - 2) ................as desired
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