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Homework Help: Cant simplify?

  1. Dec 27, 2006 #1
    the main problem im having is i cant figure out how they simplified this equation. Ill try and use latex but its my first time so i dont know if it will work.
    \int(1-\frac{1}{3(x+1)}+\frac{(x-2)}{3(x^2-x+1)}) dx
    and somehow this simplifies to this
    x- (1/3)ln(abs(x+1)) +(1/6) \int \frac{(2x-1)}{(x^2-x+1)} dx - (1/2) \int \frac{(dx)}{(x-.5)^2+(3/4)}
    i understand how to get the x-(1/3)ln(2x-1) but i dont get how they seperated the other two terms for integration. or how they pulled out (1/6)
    Last edited: Dec 27, 2006
  2. jcsd
  3. Dec 27, 2006 #2
    Given [tex]
    \int\frac{(x-2)}{3(x^2-x+1)}\, dx
    [/tex], multiply the numerator and denominator by 2 and rewrite the numerator as (2x-1)-3.
  4. Dec 27, 2006 #3


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    Ok, well the third term in the integrand of the original question is [tex]
    \frac{(x-2)}{3(x^2-x+1)}=\frac{(2x-4)}{6(x^2-x+1)}=\frac{(2x-1)}{6(x^2-x+1)}-\frac{3}{6(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x^2-x+1)}=\frac{1}{6}\frac{(2x-1)}{(x^2-x+1)}-\frac{1}{2}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} [/tex] where the last step in the denominator of the second term is done by completing the square.
  5. Dec 27, 2006 #4
    ok i understand what is going on now, but why do we chose to multiply the numerator and denominator by 2, and why do we chose 2 rather than any other number.
  6. Dec 27, 2006 #5
    Multiplying by 2 allows the fraction to be rewritten so that part of it is in the form du/u, which can be integrated to to get ln(abs(u))

  7. Dec 28, 2006 #6
    Why don't you tell us what you arrive at after using some other number/method? If it's more effective, we'll ditch the earlier one. :smile:

    At an glance, you might think that this is solvable by the substitution u = x2- x +1, but once you do it, you'll realise where the 2 comes from.
    Last edited: Dec 28, 2006
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