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Integral using partial fractions

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. The attempt at a solution

    Having difficulties with this one.
    I managed to break it down to two partial fractions, being [itex]x-1[/itex] and [itex]x^2+x+1[/itex] but couldn't make anything out of it.
  2. jcsd
  3. Oct 29, 2009 #2


    Staff: Mentor

    Yes, x3 - 1 = (x - 1)(x2 + x + 1), but what did you get for your partial fractions? They should look like this:
    [tex]\frac{A}{x - 1}~+~\frac{Bx + C}{x^2 + x + 1}[/tex]
    You should have/will need to find the values for A, B, and C.
  4. Oct 29, 2009 #3
    Sorry, forgot.

    I ended up having this:

    [tex]\frac{1}{3} \int \left(\frac{1-x}{x^2+x+1} + \frac{1}{x-1} \right)dx[/tex]

    the right part is easy, it's ln|x-1|, but the left part is what's bothering me. Tried completing the square, separating it into 2 different integrals, nothing helps.
  5. Oct 29, 2009 #4


    Staff: Mentor

    [tex] \left(\frac{1-x}{x^2+x+1} \right)[/tex]
    [tex]=~(-1/2)~\frac{(2x - 1) - 1}{x^2 + x + 1}[/tex]
    Now split this up into two fractions, and integrate each. The first one you can solve by substition, with u = x2 + x + 1.

    The second one is a little harder, as you will have to complete the square to get the denominator in the form (x + <something>)2 + <something else>. The idea here is that
    [tex]\int \frac{du}{u^2 + a^2}~=~(1/a)tan^{-1}(u/a) + C[/tex]
  6. Oct 30, 2009 #5
    You had a small mistake here, it should be (2x+1)-3 and not (2x-1)-1.

    But thanks! It helped me solve it.
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