## integral dx√(x2 - x + 1) from 0 to 1

dx√(x2 - x + 1) from 0 to 1
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 Quote by haranguyen dx√(x2 - x + 1) from 0 to 1
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.
 I will write this in a rather formal notation: $$\int_{0}^{1}\sqrt{x^2-x+1}\,dx$$ 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.

## integral dx√(x2 - x + 1) from 0 to 1

I just entered that integral in "the integrator " Can someone give an analysis of why
this integral results in such a complex evaluation ?
 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: $$\int \sec(x)dx$$ $$= \int \sec(x)\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx$$ $$= \int \frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}dx$$ Substituting $u=\sec(x)+\tan(x)$, we get $$= \int \frac{1}{u}du$$ $$= \log|u|=\log|\sec(x)+\tan(x)|+C$$ 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.

Recognitions:
Homework Help
 You may try one of the Euler substitutions: $$\sqrt{x^{2} - x + 1} = x + t$$ Then: $$x^{2} - x + 1 = x^{2} + 2 x t + t^{2}$$ $$x (2 t + 1) = 1 - t^{2}$$ 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.