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Integration by parts

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

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




    3. The attempt at a solution
    Well, this problem is hard, so I thought to use u = x3ex2
    so du = x2ex2(3+2x2) dx
    and dv = (x2+1)-2 then v = -2(x2+1)-1 Please check v though to make sure my algebra is right.

    so then using the by parts formula:
    [tex] \int \frac{x^3e^{x^2}}{x^2+1} = \frac{x^3e^{2x}}{2(x^2+1)} - \int \frac{x^2e^{x^2}(3+2x^2)}{2(x^2 +1)} [/tex] But where do I go from here?
     
  2. jcsd
  3. Jan 28, 2010 #2

    Mark44

    Staff: Mentor

    I wouldl start with u = x2/(x2 + 1)2 and dv = xex2. No guarantee that this will get you anywhere, but a useful strategy is to make dv the most complicated expression that you can integrate.
     
  4. Jan 28, 2010 #3
    Notice the fact that you have x^3 on the top. Try a u-sub with u = (1 + x^2). After that, use parts cleverly and you'll get your answer.
     
  5. Jan 28, 2010 #4

    Dick

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    Homework Helper

    This is one of those cases where there are several plausible alternatives for picking the parts, but only one works. You just have to mess around. Hint: if you pick u=x^2*e^(x^2) then du=(2xe^(x^2)+2x^3e^(x^2))dx=2(1+x^2)*e^(x^2)*x*dx. Notice the (1+x^2) factor in du. You need that.
     
  6. Jan 28, 2010 #5

    LCKurtz

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    Gold Member

    You don't get that v for your choice of dv. You probably meant

    [tex]\frac{(x^2+1)^{-1}}{-1}[/tex]

    but that's still wrong because you would need a 2x in the in the numerator of dv.
     
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