Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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].
    [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


    User Avatar

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook