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Thread: a baffling integral

  1. #1
    niz
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    a baffling integral

    I am baffled in solving the following integration problem. Please help!!

    ∫ 1 / (x^6 1) dx

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by niz View Post
    I am baffled in solving the following integration problem. Please help!!

    ∫ 1 / (x^6 – 1) dx

    Thanks in advance.
    Are you familiar with the method of partial fractions? Notice that the denominator is the difference of two squares $\displaystyle (x^3-1)(x^3+1)$. Now use the formulae for the cubes. You can write that out as $\displaystyle (x-1)(x^2+x+1)(x+1)(x^2-x+1)$

    You should be able to apply the method of partial fractions for the case when the denominator contains irreducible quadratic factors.
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  3. #3
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    Here's my method , it is for those who HATE the method of partial fraction ( but i am not the one )

    $\displaystyle \frac{1}{x^6- 1} = \frac{ x^2 - ( x^2 -1 )}{ x^6 - 1} $


    $\displaystyle = \frac{x^2}{x^6-1} - \frac{ x^2 - 1}{ ( x^2 - 1 )( x^4 + x^2 + 1 )}$

    $\displaystyle = \frac{x^2}{x^6-1} - \frac{1}{ x^4 + x^2 + 1} $


    the integral $\displaystyle \int \frac{dx}{x^6 - 1} = \int \frac{x^2 dx}{x^6-1} - \int \frac{dx}{ x^4 + x^2 + 1}$


    the first one we just need to substitute $\displaystyle t = x^3 $ and finally obtain

    $\displaystyle \frac{1}{6}\ln{ \left( \frac{ x^3 - 1}{x^3 + 1} \right ) }$


    you may think that we may need to apply partial fraction to solve the second part but if we consider


    $\displaystyle \int \frac{ x^2 + 1}{ x^4 + x^2 + 1}~dx $


    Divide the numerator and the denominator by $\displaystyle x^2 $


    $\displaystyle = \int \frac{ 1 + 1/x^2}{ x^2 + 1 + 1/x^2 } ~dx $

    $\displaystyle = \int \frac{ 1 + 1/x^2}{ \left ( x - \frac{1}{x} \right )^2 + 3 }$


    then substitute $\displaystyle x- \frac{1}{x} = t , (1 + 1/x^2 )dx = dt$

    the integral becomes

    $\displaystyle \int \frac{dt}{ t^2 + 3 } = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) $


    now consider


    $\displaystyle \int \frac{x^2 - 1 }{ x^4 + x^2 + 1}~dx $

    do the same thing above but sub. $\displaystyle x + 1/x = t $ this time , we can get

    $\displaystyle = \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } $


    we have
    $\displaystyle \int \frac{ x^2 - 1}{ x^4 + x^2 + 1}~dx = \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } $ $\displaystyle (1)$

    and

    $\displaystyle \int \frac{ x^2 + 1}{ x^4 + x^2 + 1}~dx = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } )$ $\displaystyle (2) $


    $\displaystyle (2) - (1) $ ,

    $\displaystyle \int \frac{dx}{ x^4 + x^2 + 1} = \frac{1}{2}[ \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) - \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } ] $


    therefore ,


    $\displaystyle \int \frac{dx}{x^6- 1} = \frac{1}{6}\ln{ \left( \frac{ x^3 - 1}{x^3 + 1} \right ) } - \frac{1}{2}[ \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x^2-1}{\sqrt{3} x } ) - \frac{1}{2} \ln{ \left( \frac{ x^2 - x +1}{x^2 + x + 1} \right ) } ] + C $
    Last edited by simplependulum; Dec 21st 2009 at 04:18 AM.
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  4. #4
    niz
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    Thanks

    Thanks adkinsjr and simplependulum!! Actually I was hoping for some kind of trigonometric substitution to manipulate the integrand. Like for (x^2 -1) substituting x= secy etc.. I guess there are none such solutions right? Anyway, thanks guys.. Thanks simplependulum for the smart manipulation of the integrand..that was clever.
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