1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper
    Gold Member

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