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Integration dont think this is what im meant to do

  1. Jan 19, 2009 #1
    integration,,, dont think this is what im meant to do

    in this integration problem, looks simple, but i must be missing something

    [tex]\int[/tex][tex]\frac{x-5}{x2-2x+2}[/tex]

    how do i deal with a propper fraction where i cannot simplify the denominator
     
  2. jcsd
  3. Jan 19, 2009 #2
    Re: integration,,, dont think this is what im meant to do

    (x-5)/(x^2-2x+2)
    The roots of x^2-2x+2 are x1,2=1+-sqrt(4-8)=1+-2i
    i.e (x-5)/((x-1+2i)(x-1-2i))=A/(x-1+2i)+B/(x-1-2i)

    Do you want me to stop here, or do you need another advice on how to procceed?
     
  4. Jan 19, 2009 #3

    HallsofIvy

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    Re: integration,,, dont think this is what im meant to do

    I'm not at all clear what loop quantum gravity is trying to do!

    Completing the square in the denominator gives [itex]x^2+ 2x+ 1 +1=(x+1)^2+ 1[/itex]
    Substitute u= x+ 2 so x= u- 2 and du= dx, x- 5= u- 7. The integral becomes
    [tex]\int\frac{u- 7}{u^2+1}du= \int \frac{u}{u^2+1}du+ \int \frac{1}{u^2+1}du[/tex]
    The first can be done with the second substitution [itex]v= x^2+ 1[/itex] and the second is an arctangent.
     
  5. Jan 19, 2009 #4
    Re: integration,,, dont think this is what im meant to do

    I am using partial fraction decomposition here, i.e there'a a thereoem that
    if P(x)/Q(x) and degQ(x)>=deg(P(x)), and Q(x)=(x-x0)^m1*....(x-xn)^m_n, then you can write it as: P(x)/Q(x)=A1/(x-x0)^m1+....+An/(x-xn)^m_n, are my intentions now clearer?
     
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