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Homework Help: Need help with Integration

  1. Sep 19, 2006 #1
    Integral In(x^2-x+2)dx

    First I use the intergration by parts, let u=In(x^2-x+2), du=(2x-1)dx/(x^2-x+2), dv=dx, v=x. Then it equals to xIn(x^2-x+2)-integral (x(2x-1))/(x^2-x+2) dx
    Then by using long division I get integral (2+(x-4)/(x^2-x+2))dx..but at the end I have no idea how to integral x/(x^2-x+2).......please help with this and thanks a lot.
     
  2. jcsd
  3. Sep 19, 2006 #2
    What does In(x^2-x+2)dx mean?
    Is it ( 1/(x^2 -x +2) )dx ?
     
  4. Sep 19, 2006 #3
    ln...my bad...
     
  5. Sep 19, 2006 #4

    StatusX

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    To integrate x/(x^2-x+2), first use substitution to turn this into the integral of C/(x^2-x+2) (ie, write the numerator as 1/2 (2x-1) + 1/2 and take u as the denominator), then factor the numerator as (ax+b)^2+c^2, and finally use the fact that the integral of 1/(x^2+1) is arctan(x).
     
  6. Sep 19, 2006 #5
    sorry, I don't get how "To integrate x/(x^2-x+2), first use substitution to turn this into the integral of C/(x^2-x+2) (ie, write the numerator as 1/2 (2x-1) + 1/2 and take u as the denominator)" works...

    Do you mean let x=(1/2)(2x-1)+1/2, then it will be 1/2 integral 2x/(x^2-x+2) dx, then use U substitution u=x^2 then it becomes integral (u-sqrtu+2)^(-1) du?

    I'm lost....
     
  7. Sep 19, 2006 #6

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    Yes, sorry, that wasn't very clear. I meant, write:

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

    The first term can be integrated by substitution, so you're left with the second term to integrate. That's what I meant by "turn it into the integral of C/(x^2-x+2)". Do you understand what to do from here?
     
  8. Sep 19, 2006 #7
    yes. Thanks so much!
     
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