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Integral (x e^-3x dx)

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\int x e^-3x dx[/tex]

    2. Relevant equations

    [tex]\int f(x)g'(x) = f(x)g(x) - \int f'(x) g(x)[/tex]

    Integration by substitution not allowed

    3. The attempt at a solution
    [tex] f(x) = x, f'(x) = 1, g'(x) = e^{-3x}, g(x) = \int e^{-3x} dx = -\frac{1}{3}e^{-3x}[/tex]
    [tex]\int x e^{-3x} dx = x(-\frac{1}{3})e^{-3x} - \int - \frac{1}{3} e^{-3x} dx = [/tex]
    [tex]= -\frac{1}{3} x e^{-3x} + \frac{1}{3} \int e^{-3x} dx =-\frac{1}{3} x e^{-3x} -\frac{1}{9} e^{-3x} + C[/tex]

    Which is incorrect. I'm not sure how to integrate e^(-3x) properly.
  2. jcsd
  3. Jan 10, 2012 #2
    so you tried integration by parts. what is the derivative of e^(-3x). And then how would i integrate this.
  4. Jan 10, 2012 #3


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    Why do you think this is incorrect ?
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