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Math Help - quite difficult integral

  1. #1
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    quite difficult integral

    Hi

    Evaluate:

    \int \frac{1}{\sqrt{x(x^2 +1)}}
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  2. #2
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  3. #3
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    I've 2 different solutions:

    1) This

    http://integrals.wolfram.com/index.jsp?expr=1%2F(x*Sqrt[x^2+%2B1])

    2) And this (provided by my professor)

    -\ln (x^-1 + \sqrt{x^-2 + 1})
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  4. #4
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    Use the substitution x=tan(\theta) ..
    to get (with some trig-identities):
    \sqrt{2} \int \frac{d\theta}{\sqrt{sin(2\theta})}

    Integration by parts maybe?
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  5. #5
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    Quote Originally Posted by Shock View Post
    I've 2 different solutions:

    1) This

    http://integrals.wolfram.com/index.jsp?expr=1%2F(x*Sqrt[x^2+%2B1])

    2) And this (provided by my professor)

    -\ln (x^-1 + \sqrt{x^-2 + 1})
    Wolfram says otherwise:
    http://www.wolframalpha.com/input/?i=differentiate+-ln%28%281%2Fx%29+%2B+sqrt%281+%2B+x^%28-2%29%29%29

    ie. \frac{d}{dx} \left(-\ln  \left(x^{-1} + \sqrt{x^{-2} + 1}\right) \right) = \frac{1}{x \sqrt{1 + x^2}}.. Which is close, but not the same :P
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  6. #6
    MHF Contributor chisigma's Avatar
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    Probably an elementary primitive of the function doesn't exist, so that we can try to obtain at least a series expansion of it. Let's start with...

    (1+x^{2})^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \binom {-\frac{1}{2}}{n} x^{2n} (1)

    ... where |x|<1 and...

     \binom {-\frac{1}{2}}{n} = \frac{(-\frac{1}{2})\cdot (-\frac{3}{2})\dots (-n + \frac{1}{2})}{n!} (2)

    ... and, performing integration, we obtain...

    \int \frac{dx}{\sqrt{x}\cdot \sqrt{1+x^{2}}}= \sqrt{x}\cdot \sum_{n=0}^{\infty} \binom {-\frac{1}{2}}{n} \frac{x^{2n}}{2n+\frac{1}{2}} + c (3)

    Kind regards

    \chi \sigma
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