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Integrating (using partial fractions) Apostol Section 6.25 #25

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\frac{4x^5-1}{(x^5+x+1)^2}dx[/tex]


    2. Relevant equations
    This is in the section on Partial Fractions. The main idea in this section was that you get the integral down to a sum integrals of the following forms:
    [tex]\int\frac{dx}{(x+a)^n} , \int \frac{x dx}{(x^2 + bx + c)^m} , \int \frac{dx}{(x^2+bx+c)^m}[/tex]


    3. The attempt at a solution
    The basic approach for most of these was to just use partial fractions by factoring the denominator and algebraically breaking down the result. I factored the denominator to

    [tex]\left(x^2+x+1\right)^2\left(x^3-x^2+1\right)^2[/tex]

    The term on the right could be factored again, but it doesn't look promising.


    I sense I should be using a different approach with this problem, but I'm not sure what.

    Please just give me a hint, if possible.
     
  2. jcsd
  3. Sep 2, 2011 #2

    micromass

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    Uuh, here's a way to solve it without using partial fractions. Try to write
    [tex]\frac{4x^5-1}{(x^5+x+1)^2}[/tex]

    in the form of

    [tex]\frac{f^\prime g-g^\prime f}{g^2}[/tex]

    Try to determine f here.


    But how to do it with partial fractions?? Well, the only way I see is to write

    [tex]x^3-x^2+1=(x+\alpha)(x^2-\frac{x}{\alpha^2}+\frac{1}{\alpha})[/tex]

    and proceed formally. The hope is that the [itex]\alpha[/itex] will cancel itself in the end (which it will here). But this is likely to become very difficult.
     
  4. Sep 2, 2011 #3
    The quotient rule idea is what I came to after seeing the answer, and after I saw it I couldn't un-see it. Many of the other problems in this section have been extremely cumbersome arithmetically for me, so it's very possible that the [itex]\alpha[/itex] method would be what Apostol had in mind.
     
  5. Sep 2, 2011 #4

    SammyS

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    It makes sense to assume that f(x) is a polynomial & of course g(x) = x5 + x + 1. After some consideration you can convince yourself that f(x) is of the form, f(x) = ax + b.

    WolframAlpha gets the surprisingly uncomplicated answer, but I couldn't get it to show the steps.
     
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