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Calc II - Integration of Partial Fractions

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi everyone, here is a new partial fractions question I just cannot understand:

    [tex]\int[/tex][tex]\frac{x^{3}}{x^{3}+1}[/tex]dx


    2. Relevant equations

    Partial Fractions, difference of perfect cubes, polynomial long division

    3. The attempt at a solution

    [tex]\int[/tex][tex]\frac{x^{3}}{x^{3}+1}[/tex] dx

    [tex]\int[/tex]1 dx + [tex]\int[/tex][tex]\frac{-1}{x^{3}+1}[/tex] dx

    x + [tex]\int[/tex][tex]\frac{-1}{(x+1)(x^{2}-x+1)}[/tex] dx

    [tex]\frac{A}{x+1}[/tex] + [tex]\frac{B}{(x^{2}-x+1)}[/tex]

    A(x[tex]^{2}[/tex]-x+1) + B(x+1) = - 1

    If I use coefficients:
    x[tex]^{2}[/tex]: 0=A
    x: 0 = B-A
    : -1 = A+B

    These don't add up! Also, using critical values leads me to a similar problem. What am I doing wrong in my strategy of attacking these partial fractions??

    Thank you for your time and help, you all are so wonderful!
     
  2. jcsd
  3. Sep 21, 2008 #2
    [tex]\frac{A}{x+1}+\frac{B}{(x^{2}-x+1)}[/tex]

    Remember that when you decompose the fraction, the numerator must be a general polynomial with degree one less than that of the denominator. What I'm getting at is that the numerator in your "B"-term is wrong. What should it be instead?
     
  4. Sep 21, 2008 #3
    Hmm, I don't quite understand what you are saying. I thought I was just supposed to multiply the numerator by the denominator and cancel. Is there another concept I am missing?
     
  5. Sep 21, 2008 #4

    statdad

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    If the denominator is linear, the numerator must be a constant. If the denominator is an irreducible quadratic, the numerator should be a ********** (hint: not a constant)
     
  6. Sep 21, 2008 #5

    gabbagabbahey

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    The numerator of each term must be a polynomial of degree one less than the degree of the denominator. [tex]x^2-x+1[/tex] is a 2nd degree polynomial, so the numerator for that term should be a first degree polynomial; [tex]Bx+C[/tex].
     
  7. Sep 21, 2008 #6
    In your second term, you just put a B in the numerator. But that is a zeroth order polynomial when you have a second order polynomial in the denominator. So instead of just B, you need to put in an arbitrary first order polynomial. What is another name for a first order polynomial?
     
  8. Sep 21, 2008 #7
    Right gabbagabbahey. Notice that the largest exponent in the denominator is 2. So the largest exponent in the numerator should be 1 (i.e. one less than 2).
     
  9. Sep 21, 2008 #8
    So it should be Bx?
     
  10. Sep 21, 2008 #9
    ohh I see gabbagabbahey, I should have a Bx and a separate C over the same denominator?
     
  11. Sep 21, 2008 #10

    gabbagabbahey

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    You should have:
    [tex]\frac{A}{x+1}+\frac{Bx+C}{(x^{2}-x+1)} [/tex]
     
  12. Sep 21, 2008 #11
    Yes. By "general" or "arbitrary" we mean that you have to include all the terms. e.g. for a second order polynomial you'd have
    [tex]Ax^2+Bx+C[/tex].
    You must include all the lower order terms too (i.e. the 2nd (A), 1st (B) and 0th (C) terms).
     
  13. Sep 21, 2008 #12
    Ok! I got it now! Thank you both! Hopefully I'll never have to post about partial fractions again :)
     
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