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Does it?

  1. Nov 12, 2007 #1
    Can the integral be found?

    Does the integral [tex]\int \sqrt{u^{4}+1} dx[/tex] have a solution?
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2


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    What exactly are you asking? In the integral you give, you have a function of u integrated with respect to x! Is u some function of x or is that simply a typo?
    If you meant [itex]\int\sqrt{x^4+1}dx[/itex], then since [itex]\sqrt{x^4+1}[/itex] is continuous, yes, there certainly exist a function having that as its derivative- it has an anti-derivative.

    If, however, you are asking whether that anti-derivative can be written in terms of "elementary functions", no it cannot.
  4. Nov 12, 2007 #3
    I'm sorry, yes that is what I meant. I thought not. Thanks a lot!
  5. Nov 12, 2007 #4
    Elliptic Functions.
  6. Nov 13, 2007 #5
    Actually, I encountered the integral in a contest. The question was:

    [tex]\frac{d^{2}}{(dx)^{2}}\int^{x}_{0}\left(\int^{sin t}_{1}\sqrt{u^{4}+1} du\right) dt}[/tex]

    I wasn't so sure how to solve it, but what I did was:
    (I'm not even sure this is proper use of the fundamental theorem of the Calculus...)

    [tex]=\frac{d^{2}}{(dx)^{2}} \int \left(\int^{sin x}_{1}\sqrt{u^{4}+1}du+\int^{sin 0}_{1}\sqrt{u^{4}+1}du\right)dx[/tex]

    [tex]=\frac{d}{dx} \left(\int^{sin x}_{1}\sqrt{u^{4}+1}du+\int^{sin 0}_{1}\sqrt{u^{4}+1}du\right)dx[/tex]

    [tex]=\frac{d}{dx}\left(\int \sqrt{sin^{4}x+1}dx-\int \sqrt{2}dx - \left(\int dx-\int \sqrt{2}dx\right) \right)[/tex]

    [tex]=\frac{d}{dx}\left(\int \sqrt{sin^{4}x+1}dx-\int dx\right)[/tex]

    But they said the answer was [tex]cos x \sqrt{sin^{4}x+1}[/tex]

    Could anyone please tell me what I did wrong? Thanks
  7. Nov 13, 2007 #6
    I think when you replaced the u with sinx you forgot to also do the chain rule. That would get you the cosx.
  8. Nov 13, 2007 #7
    oh yeah... makes sense. We were never really taught this stuff in class, (I don't think they deemed it necessary for IT majors to learn theory.) Thanks a lot!
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