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Exponent integral problem

  1. May 2, 2005 #1
    i solved ODE with Lagrange method and got stuck with integral

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

    i couldnt solve it with any method but combined and got the answer that it is

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

    the problem is that i want to do it with proper method, and show how it comes out. My mentor said that these types of integrals are "freaky" with little twist:smile:

    Anyway, i dont mind some advice
     
    Last edited: May 2, 2005
  2. jcsd
  3. May 2, 2005 #2

    dextercioby

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    U can do it using part integration.The derivative of the exponent is [itex] x+1[/itex] which u can obtain writing [itex] x^{2}=x(x+1)-x [/itex]


    Daniel.
     
  4. May 3, 2005 #3
    i dont get how that helps me
     
  5. May 3, 2005 #4

    dextercioby

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    It does

    [tex] \int x^{2}e^{\frac{1}{2}x^{2}+x} \ dx=\int \left(x^{2}+x-x\right) e^{\frac{1}{2}x^{2}+x} \ dx=\int x(x+1)e^{\frac{1}{2}x^{2}+x} \ dx-\int x e^{\frac{1}{2}x^{2}+x} \ dx[/tex]
    [tex]=xe^{\frac{1}{2}x^{2}+x}-\int e^{\frac{1}{2}x^{2}+x} \ dx-\int x e^{\frac{1}{2}x^{2}+x} \ dx [/tex]
    [tex]=x e^{\frac{1}{2}x^{2}+x}-\int (x+1)e^{\frac{1}{2}x^{2}+x} \ dx=x e^{\frac{1}{2}x^{2}+x}-e^{\frac{1}{2}x^{2}+x}+C [/tex]


    Daniel.
     
  6. May 4, 2005 #5
    wow, that is very clever, at first i looked ur answer and couldn`t get one line but then i realized that the key moment was to crack int x(x+1)*e^...dx with simple method by taking u=x and dv=(x+1)*e^....dx and now i realize the beauty. So thank u very much for helping me on this one.
     
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