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Math Help - another difficult integral

  1. #1
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    another difficult integral

    Hi everyone,

    Could someone please tell me how I would solve the integral 1 to infinity x/sq.1+x^6?

    I did separation by parts?

    u=sq. 1+x^6
    dv=xdx
    du=1/2(1+x^6)(6x^5)dx
    v=x^2/2

    Would this work?

    Thank you very much
    Last edited by chocolatelover; October 7th 2007 at 09:36 AM.
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  2. #2
    Senior Member polymerase's Avatar
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    that would work, you would get

    \dfrac{x^2}{\sqrt{1+x^6}}-\dfrac{3}{7}\int x^7\,dx
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  3. #3
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    Careful there, integration by parts is not well applied.
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  4. #4
    Senior Member polymerase's Avatar
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    sry...man i did wrong i was reading your question n doing my own at the same time....
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Could someone please tell me how I would solve the integral 1 to infinity x/sq.1+x^6?

    I did separation by parts?

    u=sq. 1+x^6
    dv=xdx
    du=6x^5ddx
    v=x^2/2

    Would this work?

    Thank you very much
    \frac d{dx} \sqrt{1 + x^6} is NOT 6x^5
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  6. #6
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    I got lim as t goes to infinity [sq. 1+x^6(x^2)/(2)-1/4int. 6x^12+6x^7]1 to t

    is that correct?

    Thanks
    Last edited by chocolatelover; October 7th 2007 at 10:44 AM.
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  7. #7
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    Then I got lim as t goes to infinity [sq. 1+t^6(t^2/2)-1/4((6t^13/13+6(t)^8/8]-[sq.1+1^6(1^2)/(2)-1/4(6(1)^13/13+6(1)^8/8]

    Is that correct?

    I wouldn't have use L'Hopital's rule, would I? It would just be infinity, right?

    Thank you
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  8. #8
    Eater of Worlds
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    Just for your benefit, I don't believe this is integrable by elementary means.
    Therefore, parts will not work too well.

    \int_{1}^{\infty}\frac{x}{\sqrt{1+x^{6}}}dx

    I ran it through Maple and got 0.9473799840

    For the indefinite, I got: \frac{x^{2}}{2}hypergeom\left([\frac{1}{3},\frac{1}{2}],[\frac{4}{3}],-x^{6}\right)


    Just a thought to throw out there.
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  9. #9
    Senior Member polymerase's Avatar
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    did you check to see if there is actually an answer because i dont think this function can even be integrated....
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    Just for your benefit, I don't believe this is integrable by elementary means.
    Therefore, parts will not work too well.

    \int_{1}^{\infty}\frac{x}{\sqrt{1+x^{6}}}dx

    I ran it through Maple and got 0.9473799840

    For the indefinite, I got: \frac{x^{2}}{2}hypergeom\left([\frac{1}{3},\frac{1}{2}],[\frac{4}{3}],-x^{6}\right)


    Just a thought to throw out there.
    Well, seeing as the indefinite integral has a solution the definite integral ought to as well. The hypergeometric function is a series solution so I wonder if chocolatelover should just derive a Taylor series for the integrand and integrate term by term? Of course then you have to worry about series convergence, etc, so it wouldn't be a walk in the park.

    -Dan
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  11. #11
    Eater of Worlds
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    As I stated previously, the solution is 0.9473799840.

    Of course, I arrived at this via Maple 10.

    Here's a graph using Simpson's rule and 100 partitions. I couldn't use 1 to infinity, so I just used 1 to 100. The graph gets infinitely close to the x-axis, so it's not that bad of an estimate
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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