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Math Help - integration by parts

  1. #1
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    integration by parts

    integral(x^2)(e^-x)dx

    the answer given was (-x^2)(2e^-x - e^-x)+c

    during one of the steps i have (-x^2) + 2[-x(e^-x)+integral(e^-x)dx]

    so when you multiply or distribute that second part by the 2 shouldnt you end up with (-x^2)(2e^-x - 2e^-x)+c ?
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  2. #2
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    Hello, jeph!

    You and your book are dropping off parts of the problem . . .


    (x)e^{-x} dx
    By parts:

    . . . u = x . . . . dv = e^{-x} dx
    . . du = 2x dx . . . v = -e^{-x}

    We have: .-(x)e^{-x} + 2
    xe^{-x} dx


    By parts again . . .

    . . . u = x . . . . dv = e^{-x} dx
    . . du = dx . . . . v = -e^{-x}

    We have: .-(x)e^{-x} + 2[-xe^{-x} + ∫e^{-x} dx]

    . . . . . .= .-(x)e^{-x} + 2[-xe^{-x} - e^{-x}] + C

    . . . . . .= .-(x)e^{-x} - 2xe^{-x} - 2e^{-x} + C

    . . . . . .= .-e^{-x} (x + 2x + 2) + C

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