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∫1/ (x^3-1)

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫1/(x^3-1)


    2. Relevant equations
    A/(x-1) + (Bx+C)/(x^2+x+1)
    1=A(x^2+x+1) + (Bx+C)(x-1)

    3. The attempt at a solution
    I've solved for all the variables in this...
    A= 1/3
    B= -1/3
    C= -2/3

    which gets me:
    1/3*Ln|x-1| + ∫(-(x/3)-(2/3))/(x^2+x+1)

    I can't figure out how to integrate the second half. Maybe i'm not using the right u (u-sub)? There's probably something really simple i'm not seeing.
     
  2. jcsd
  3. Jun 15, 2012 #2
    Re: ∫1/(x^3-1)

    What if you had 1/(ax^2+b)? Then you would use trig sub right? If you complete the square, then you'll be closer to this form.
     
  4. Jun 15, 2012 #3
    Re: ∫1/(x^3-1)

    Hi arishorts! Welcome to PF!! :biggrin:

    Try substituting [tex]u = x^2 + x + 1[/tex]
     
  5. Jun 15, 2012 #4
    Re: ∫1/(x^3-1)

    You already know this... but i get du=2x+1 dx, but the numerator is x+2. is there a transition i'm missing?
     
  6. Jun 15, 2012 #5
    Re: ∫1/(x^3-1)

    sorry, i didn't notice the x factor in the numerator, you should integrate that part spearately, you can use the easier u substitution.
     
  7. Jun 15, 2012 #6
    Re: ∫1/(x^3-1)

    I'm trying this right now, don't see why it shouldn't work. Is there not a simpler way?


    Edit:

    This is what i'm doing right now:

    -(1/3)∫ (x+2) / ((3/4)-(x+(1/2))^2)
    Then i'd trig sub...
     
  8. Jun 15, 2012 #7
    Re: ∫1/(x^3-1)

    Write it as

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

    And then complete the square.
     
  9. Jun 15, 2012 #8
    Re: ∫1/(x^3-1)

    du/2=(x+1/2)dx, then for the leftover (a constant in the numerator), complete the square then use trig sub.
     
  10. Jun 15, 2012 #9
    Re: ∫1/(x^3-1)

    no simpler way, that's why they teach trig sub

    EDIT: Infinitum's last post looks good, use simpler u-sub on term on left, trig sub on right.
     
  11. Jun 15, 2012 #10
    Re: ∫1/(x^3-1)

    how can you do that to the numerator, but not the denominator?
     
  12. Jun 15, 2012 #11
    Re: ∫1/(x^3-1)

    Is this correct?

    [tex]\frac{a+b}{c+d} = \frac{a+b}{c} + \frac{a+b}{d}[/tex]

    Put in some values to be sure :wink:
     
  13. Jun 15, 2012 #12
    Re: ∫1/(x^3-1)

    You seem to have a very strong algebraic manipulation background, where as i don't. Been trying to work on it lately. I'm trying all methods mentioned here (trying to make sense of it all).
     
  14. Jun 15, 2012 #13
    Re: ∫1/(x^3-1)

    Infinitum actually multiplied by 2(1/2), rather than (1/2)/(1/2). You see, they distributed the 2 to the numerator, and left the 1/2 out front.
     
  15. Jun 15, 2012 #14
    Re: ∫1/(x^3-1)

    do what to the numerator?
     
  16. Jun 15, 2012 #15
    Re: ∫1/(x^3-1)

    not trying to be an ***...

    a=1, b=2, c=3, d=4

    3/7 ≠ 3/3 + 3/4
     
  17. Jun 15, 2012 #16
    Re: ∫1/(x^3-1)

    it was edited. the 1/2 wasn't there before or i didn't see it. i see it now. Let me try this.
     
  18. Jun 15, 2012 #17

    SammyS

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    Re: ∫1/(x^3-1)

    [itex]\displaystyle -\frac{1}{3}\frac{x+2}{x^2+x+1}=-\frac{1}{3}\frac{1}{2}\frac{2x+4}{x^2+x+1}[/itex]
    [itex]\displaystyle =-\frac{1}{6}\frac{2x+1+3}{x^2+x+1}[/itex]

    [itex]\displaystyle =-\frac{1}{6}\left(\frac{2x+1}{x^2+x+1}+\frac{3}{x^2+x+1}\right)[/itex]​
     
  19. Jun 15, 2012 #18
    Re: ∫1/(x^3-1)

    Got it! Thanks guys... Give you guys +rep if i could, but there doesn't seem to be that system here.
     
  20. Jun 15, 2012 #19
    Re: ∫1/(x^3-1)

    Just got it. Thanks.
     
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