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Homework Help: Confusing Definite Integral

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Integrate [tex]\int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx[/tex]

    2. Relevant equations

    3. The attempt at a solution
    U sub: let [tex]u=x^2-x+3[/tex] Then [tex] du=2x-1[/tex] and then have to evaluate [tex]\int_{3}^{3}\sqrt{\frac{du^2}{u}}dx[/tex] But how with these limits of integration should this be 0? Not sure how to evaluate this....
  2. jcsd
  3. Mar 12, 2008 #2
    Complete the square in both numerator and denominator, then use a u-sub for the radican.
  4. Mar 12, 2008 #3


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    Science Advisor

    I see no reason to do that.

    [itex]\sqrt{x^2}= |x|[/itex] and 2x- 1 changes sign at x= 1/2. This integral is the same as
    [tex]\int_0^1 \frac{|2x-1|}{\sqrt{x^2- x+ 3}}dx= -\int_0^{1/2}\frac{2x-1}{\sqrt{x^2-x+ 3}}dx+ \int_{1/2}^1 \frac{2x-1}{\sqrt{x^2- x+ 3}}dx[/tex]
  5. Mar 12, 2008 #4
    Thank you so much. I was getting confused with changing the limits of integration back and forth but got it now, thanks.
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