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Have i integrated this correctly?

  1. Jan 26, 2009 #1
    [tex]\int[/tex](x-5)/(x2-2x+2)dx

    (x-5)/(x2-2x+2)=(x-1-4)/((x-1)2+1)

    x-1=t therefore x=t+1

    dx=x'dt=(t+1)'dt=dt


    [tex]\int[/tex](x-5)/(x2-2x+2)dx=[tex]\int[/tex](t-4)/(t2+1)dt

    =[tex]\int[/tex]t/(t2+1)dt-4[tex]\int[/tex]1/(t2+1)dt

    =0.5ln|t2+1|-4arctg(t)+c
     
  2. jcsd
  3. Jan 26, 2009 #2

    Mark44

    Staff: Mentor

    If you differentiate your answer, you should be able to get back to your original integrand.
     
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