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Math Help - Integration of 1/(1+x^4)^(1/4)

  1. #1
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    Integration of 1/(1+x^4)^(1/4)

    So I have been having trouble with finding the indefinite integral of
    1/ (1+x^4)^(1/4), by this I mean (1+x^4)^(-1/4).
    I was given a hint that says to use the substitution t^4= 1+x^(-1/4).
    Any help would be appreciated, and thanks in advance!
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  2. #2
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    Wolfram says that the solution involves the Hypergeometric Function.

    http://www.wolframalpha.com/input/?i=integral[%281+%2B+x^4%29^%28-1%2F4%29]
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  3. #3
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    I tell you

    There exists a solution , perhaps the only one , to this integral

     \int \frac{dx}{ (1 + x^n)^{\frac{1}{n}}}


    Which is substituting  1 + x^{-n} = t^n <- be cafeful

    the power of  x is this time -n not n

    so

     -n x^{-n-1}dx = n t^{n-1} dt

     dx = - x^{n+1} t^{n-1} dt


     \int \frac{dx}{ (1 + x^n)^{\frac{1}{n}}}


     \int \frac{1}{ (1 + x^n)^{\frac{1}{n}}} \cdot ( - x^{n+1} t^{n-1} dt )

     = - \int \frac{x^{n+1} t^{n-1}}{x (1 + x^{-n})^{\frac{1}{n}}} ~dt

     - \int \frac{x^{n} t^{n-1}}{ t} ~dt

     - \int \frac{t^{n-2}}{t^n - 1} ~dt

    When  n = 4


    the integral is tansformed to

     - \int \frac{ t^2 }{ t^4 - 1}~dt

     = -\frac{1}{2} \int \left [ \frac{1}{t^2 - 1} + \frac{1}{t^2 + 1} \right ]~dt
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Wolfram says that the solution involves the Hypergeometric Function.

    http://www.wolframalpha.com/input/?i=integral[%281+%2B+x^4%29^%28-1%2F4%29]
    You don't always have to believe to Wolfram.

    In the integral put x=\frac1z and it becomes -\int\frac{dz}{z\sqrt[4]{1+z^4}}, now put t^4=1+z^4 and the integral is -\int\frac{t^2}{t^4-1}\,dt which is the same as simplependulum gave.
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