By completing the square, show that $\int_{0}^{\frac{1}{2}}\frac{dx}{x^2-x+1}=\frac{\pi }{3\sqrt{3}}$

Question

By completing the square, show that $\int_{0}^{\frac{1}{2}}\frac{dx}{x^2-x+1}=\frac{\pi }{3\sqrt{3}}$. I found that $\int_{0}^{\frac{1}{2}}\frac{dx}{x^2-x+1}$ equals to $\int_{0}^{\frac{1}{2}}\frac{x+1}{x^3+1}dx$ so it becomes $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{3n+1}}{(3n+1)}+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{3n+2}}{(3n+2)}$, but can't go any further.

Answer 1

Complete the square in denominator per your own hint. \begin{align} I &= \int_0^{\frac 12} \frac {dx}{x^2-x+1} = \int_0^{\frac 12} \frac {dx}{\left ( x^2 -x + \frac 14\right ) + \frac 34} = \int_0^{\frac 12} \frac {dx}{\left( x- \frac 12\right)^2 + \frac 34} = \\ &= \int_{-\frac 12}^0 \frac {dt}{t^2 + \frac 34} = \left . \frac 2{\sqrt 3} \arctan \frac {2t}{\sqrt 3} \right |_{-\frac 12}^0 = ??? \end{align} Can you take it from here?

Answer 2

It goes like this: $x^2 - x + 1 = (x - \dfrac{1}{2})^2 + (\sqrt{3}/2)^2$, and you can take: $u = \sqrt{3}/2\cdot tan(x -\frac{1}{2})$ to transform it into an $tan^{-1}$ anti derivative form.