1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    StatusX

    User Avatar
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Need help with Integration
Loading...