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Integral dx√(x2 - x + 1) from 0 to 1

  1. Jun 18, 2012 #1
    dx√(x2 - x + 1) from 0 to 1
  2. jcsd
  3. Jun 18, 2012 #2


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    Re: integral

    Hey haranguyen and welcome to the forums.

    Can you show any work or ideas you have for tackling this problem? We don't do other people's homework for them, but we do make an effort to assist them in they provide the above.
  4. Jun 18, 2012 #3
    Re: integral

    I will write this in a rather formal notation:
    I solved this integral in about 3-4 minutes. If you have some work, please put it here. Otherwise, I will tell you where to start.
  5. Jun 18, 2012 #4


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    Re: integral

    I just entered that integral in "the integrator " Can someone give an analysis of why
    this integral results in such a complex evaluation ?
  6. Jun 18, 2012 #5
    Re: integral

    Such integrals often involve the integral of secant cubed, which involves the integral of secant. To integrate the secant function, you can go like this:

    [tex]\int \sec(x)dx[/tex]
    [tex]= \int \sec(x)\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx[/tex]
    [tex]= \int \frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}dx[/tex]
    Substituting [itex]u=\sec(x)+\tan(x)[/itex], we get
    [tex]= \int \frac{1}{u}du[/tex]
    [tex]= \log|u|=\log|\sec(x)+\tan(x)|+C[/tex]

    It shouldn't be hard to derive the integral of secant cubed from there.
    The integral of secant cubed often appears in radical integrals like this. It would be useful to memorize it.
  7. Jun 18, 2012 #6


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    Re: integral

    Easier with sinh, no?
  8. Jun 18, 2012 #7
    Re: integral

    Pretty much the same for me.
  9. Jun 19, 2012 #8


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    I did a quick graph of this function y = √x2-x+1
    And it looks like it is approaching linear:
    x=2 y= 1.73
    x=3 y= 2.64
    x=4 y= 3.6
    x=20 y= 19.5
    x=30 y= 29.6
    x=40 y=39.76
    From Wolfram : ∫ √x2-x+1 dx
    =√x2-x+1(x/2-1/4)+3/8 sinh-1(2x-1/√3)
    Not questioning the evaluation but the complexity with respect to the graph ?
  10. Jun 19, 2012 #9
    You may try one of the Euler substitutions:
    \sqrt{x^{2} - x + 1} = x + t
    x^{2} - x + 1 = x^{2} + 2 x t + t^{2}
    x (2 t + 1) = 1 - t^{2}
    Then, you can solve for x and substitute back in the original substitution to express the square root in terms of t. Also, you can differentiate to get dx. Finally, what are the limits for t?

    You should now have an integral of a rational function.
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