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Integration by parts problem

  1. Nov 29, 2004 #1
    Hi, I've actually got a problem here.

    How do I evaluate

    [tex]\int e^x^3 x^2 dx[/tex]

    I have problem when doing integration by parts of finding [tex] \int v du [/tex] since if I integrate v du, i'll get another expression which i have to integrate by parts again, and this goes on and on !


    (its meant to be e to the power x cubed by the way).
     
  2. jcsd
  3. Nov 29, 2004 #2

    dextercioby

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    It probably reads:
    [tex]\int e^{x^3} x^2 dx[/tex].
    If so,this integral is trivial and it does not require anything,not even the lousy substitution [tex] x^3 =u [/tex].
    So:
    [tex]\int e^{x^3} x^2 dx =\frac{1}{3} e^{x^3} +C [/tex].
    If it's not as i interpreted it,well,then 2 integrals by parts should simply do the trick.
    Good luck!!
     
    Last edited: Nov 29, 2004
  4. Nov 29, 2004 #3

    WaR

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    Hello
    It doesn't go on forever. You only do it twice.
    The first time [tex]u = x^2[/tex] and [tex]dv = e^{3x} dx[/tex]

    Then you get the following:
    [tex] \frac{1}{3} x^2 e^{3x} - \int \frac{1}{3} e^{3x} 2x dx[/tex]

    Now, you do integration by parts a second time
    This time [tex]u = 2x[/tex] and [tex]dv = e^{3x} dx[/tex] (remember you can pull out that 1/3)

    Then you get the following:
    [tex] \frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \int \frac{1}{3} e^{3x} 2 dx[/tex]

    The last integral doesn't need integration by parts, it's just a simple integration. You should get the following:

    [tex]\frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27}e^{3x}[/tex]


    Sorry for not using LateX. I'll get the hang of it later though :)


    Edit: Added LateX. That takes forever.
     
    Last edited: Nov 29, 2004
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