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Calculus: partial fractions

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Integration of 1/(x^2-5x+6)

    2. Relevant equations

    3. The attempt at a solution
    I know i cannot do ln|x^2-5x+6|
    I've tried some form of substitution or intergration by parts, and they don't work.
    Should I factor the bottom?
  2. jcsd
  3. Feb 6, 2008 #2
    Have you tried partial fractions, as the title suggests? To be brief, you can rewrite [tex]\frac{1}{x^2-5x+6}[/tex] as [tex]\frac{a}{x-2}+\frac{b}{x-3}[/tex] where you must find a and b.
    Last edited: Feb 6, 2008
  4. Feb 7, 2008 #3
    Okay.. I've solved the question up to a certain point. (I've decided that it's do-able without partial practions..)
    however; now I am stuck on an integration:
    i might find the integral of
    and now im not sure if i use substituion, integration by parts, or bring the bottom up to the top of the fraction for this integration?
  5. Feb 7, 2008 #4
    Ok, show us then. Maybe we can help you from there.
  6. Feb 7, 2008 #5
    So partial fractions is the way to go I think..
    I have
    A/(x-3) + B/(x-2) = )A(x-2)+B(x-3))/(x-2)(x-3)
    Ive solved for A and B and I got 1 for each of them.
    Now I need to go and integrate 1/(x-2) and 1/(x-3) right? and set that equal to (t+c) and then solve the autonomous equation as i would any?
  7. Feb 7, 2008 #6
    Yes, you would integrate it from here.
  8. Feb 7, 2008 #7


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    A=1 and B=1 doesn't work. Think about the signs again.
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