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Re-write the denominator as x*(x^6-1)=x*(x^3-1)*(x^3+1)

  1. Dec 1, 2009 #1
    Can any one help me with this [tex]\int \frac{dx}{x^7-x}[/tex]?
     
    Last edited: Dec 1, 2009
  2. jcsd
  3. Dec 1, 2009 #2

    arildno

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    Re: Integral

    1. Re-write the denominator as x*(x^6-1)=x*(x^3-1)*(x^3+1)

    2. Use polynomial division to reduce the third-degree polynomials:
    [tex](x^{3}\pm{1}):(x\pm{1})=x^{2}\mp{x}+1[/tex]

    3. See if these can be factorized any further, then use partial fractions decomposition.
     
  4. Dec 1, 2009 #3
    Re: Integral

    but it takes an hour to do it
    is there any simpler method for this?
     
  5. Dec 2, 2009 #4

    HallsofIvy

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    Re: Integral

    I guess you had better stick to trivial problems!
     
  6. Dec 3, 2009 #5
    Re: Integral

    1/(x^7-x) = 1/(x*(x^6-1))= -1/x+x^5/(x^6-1)
    int(-1/x+x^5/(x^6-1), x)=
    int(-1/x, x)=-ln(x)
    int(x^5/(x^6-1), x)=(1/6)*ln(x^6-1)
    int(1/(x^7-x), x)=-ln(x)+(1/6)*ln(x^6-1)
     
  7. Dec 3, 2009 #6

    arildno

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    Re: Integral

    I didn't see that one..

    Very good, saeed69! :smile:
     
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