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How to find the integral of x 3^(x^2)

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex] x 3^(x^2)


    2. Relevant equations

    integration by parts

    3. The attempt at a solution

    u= x dv= 3^(x^2)
    du = 1 dx v=((3^(x^2))/ln 3 ) 2x + c ??

    (x)((3^(x^2))/ln 3 ) 2x - [tex]\int[/tex] ((3^(x^2))/ln 3 ) 2x dx

    is this right??
     
  2. jcsd
  3. Feb 6, 2009 #2
    Re: Integrate

    No: what do you get when you differentiate your v? 3^(x^2) doesn't have a nice integral.

    Rather, you should recognise [tex]\int x3^{x^2} \, dx[/tex] is built for a substitution: let [tex]u = x^2[/tex].
     
  4. Feb 6, 2009 #3
    Re: Integrate

    so u=x^2
    du = 2x

    1/2 3^u du??
     
  5. Feb 6, 2009 #4
    Re: Integrate

    do you mean

    [tex]

    \frac{1}{2}\int 3^u du

    [/tex]
     
  6. Feb 6, 2009 #5
    Re: Integrate

    are u sure about this.. people are telling me that i need to integrate by parts
     
  7. Feb 6, 2009 #6
    Re: Integrate

    Unco is correct: use the substitution
     
  8. Feb 6, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Integrate

    Then stop listening to those people!
     
  9. Feb 6, 2009 #8
    Re: Integrate

    k gotcha. quick question whats the difference between integrating by parts and u substitution
     
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