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Math Help - Cube rooted integral

  1. #1
    Newbie superphysics's Avatar
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    Cube rooted integral

    I need to solve the following question:

    \int{\sqrt[3]{x^2+1}} dx

    Any help/solution will be much appreciated.
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  2. #2
    Eater of Worlds
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    Here's what Maple gave me:

    x\cdot\text{hypergeom}\left( [1/3,1/2],[3/2],-{x}^{2} \right)
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  3. #3
    Super Member wingless's Avatar
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    I think this is not a solvable integral.

    Mathematica gave:
    \frac{1}{5} x \left(3 \left(1+x^2\right)^{1/3}+2 \text{Hypergeometric2F1}\left[\frac{1}{2},\frac{2}{3},\frac{3}{2},-x^2\right]\right)
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  4. #4
    Eater of Worlds
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    It does not have a nice indefinte form. If it had limits of integration, perhaps Kriz or someone could have a clever way to do it?. Here's the graph. At first glance, one would think it'd be easy.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  5. #5
    Newbie superphysics's Avatar
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    Ok, if it's unsolvable indefinitely, can we do it in the range 0 to a?

    Some clever trick?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by superphysics View Post
    Ok, if it's unsolvable indefinitely, can we do it in the range 0 to a?

    Some clever trick?
    What galactus meant was something like from 0 to \infty. You can't put this into a form with a variable.

    You can put it in the form of an infinite series. (That's how the hypergeometric function is typically written.)

    -Dan
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