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Help with this integral?

  1. Jul 25, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int (x^{2})(3^{x^{3}}) dx[/itex]

    2. Relevant equations



    3. The attempt at a solution

    let u = x^3, du = 3x^2 dx

    [itex]\frac{1}{3}\int 3^{u}du[/itex]

    [itex]\frac{1}{3} (\frac{1}{ln 3})3^{u}[/itex]

    [itex]\frac{1}{3} (\frac{1}{ln 3})3^{x^{3}}[/itex]
     
  2. jcsd
  3. Jul 25, 2011 #2

    BruceW

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    Yep. You got it right.
     
  4. Jul 25, 2011 #3
    *Sigh* Mathway never agrees with me, or says it can't solve the problem. Wolfram also showed me something weird.

    I guess I need to stop second guessing myself when those online solvers give a weird answer.
     
  5. Jul 25, 2011 #4

    dextercioby

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    Don't forget the constant of integration, each time you use the integral sign without limits. Wolfram is 99.99% right.
     
  6. Jul 25, 2011 #5

    SammyS

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    (1/3) = 3-1

    Therefore, [itex]\displaystyle \frac{1}{3} \left(\frac{1}{\ln 3}\right)3^{x^{3}}=\left(\frac{1}{\ln 3}\right)3^{(x^{3}-1)}\,,[/itex] which is pretty much what WolframAlpha gives.
     
  7. Jul 25, 2011 #6
    I think that was wolfram's result. Now I see why.
     
  8. Jul 25, 2011 #7

    gb7nash

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    Wolfram is correct. It gives me:

    [tex]\frac{3^{x^3-1}}{\log{3}}[/tex]

    which is:

    [tex]\frac{3^{x^3}3^{-1}}{\log{3}} = \frac{1}{3}\frac{3^{x^3}}{\log{3}}[/tex]

    Sammy beat me.
     
  9. Jul 25, 2011 #8
    Whenever my calculator tells me it can't solve an integral I always try making some substitution or similar adjustments (especially trig). Sometimes the ability to see pieces of a puzzle is lacking in straight-up algorithms.
     
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