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Homework Help: Integration by parts

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Show using integration by parts that:

    [tex] \int x^3 e^x^2 dx = e^x^2 ( \frac{ x^2 -1}{ 2 }) [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Integration by parts obviously.

    [tex] \int u dv = uv - \int v du [/tex]

    Let [tex] u = x^3 [/tex] and [tex] dv = e^x^2 dx [/tex]

    [tex] \int x^3 e^x^2 dx = \frac{x^2 e^x^2}{ 2 } - \frac{3}{2} \int x e^x^2 dx [/tex]

    Now use integration by parts again on [tex] \int x e^x^2 dx [/tex]

    And I get :

    [tex] \frac{e^x^2}{ 2 } - \frac{1}{2} \int \frac{1}{x} e^x^2 dx [/tex]

    This really leaves me no closer again because I have to use integration by parts again on

    [tex] \int \frac{1}{x} e^x^2 dx [/tex]

    Any suggestions on what to do.
    Thanks for the help.
     
  2. jcsd
  3. Nov 8, 2008 #2

    Dick

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    You mean e^(x^2). Your superscript isn't coming out. On the first step, you can't pick dv=e^(x^2)*dx. You can't integrate that. I have no idea what you are doing after that. Try dv=x*e^(x^2)*dx and u=x^2 for a first step. When you get to xe^(x^2), don't do parts again. Do it by an easy u-substitution (the same one you used to integrate dv).
     
  4. Nov 8, 2008 #3

    gabbagabbahey

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    I don't think this is a very good choice for your [itex]u[/itex] and [itex]dv[/itex], because [tex]v=\int dv=\int e^{x^2} dx [/tex] is not [itex]e^{x^2}[/itex]....try a substitution of the form [itex]w=x^2[/itex] before applying integration by parts :wink:
     
  5. Nov 8, 2008 #4
    Cool, I showed it.
    Thanks for the help!
     
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