[tex]\displaystyle \int_{0}^1 \dfrac{1}{x^2+x+1}\, dx = \int_{0}^1\dfrac{1}{\Big(x+\dfrac{1}{2}\Big)^2+\dfrac{3}{4}}\, dx = \\ \\ = \int_{0}^1\dfrac{\Big(x+\dfrac{1}{2}\Big)'}{\Big(x+\dfrac{1}{2}\Big)^2+\Big(\dfrac{\sqrt 3}{2}\Big)^2}\, dx = \dfrac{2}{\sqrt 3}\, \mathrm{arctg}\,\left(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt 3}{2}}\right)\Bigg|_{0}^1 = \\ \\ = \dfrac{2\sqrt 3}{3}\Bigg(\,\mathrm{arctg}\,\dfrac{3}{\sqrt 3}-\, \mathrm{arctg}\, \dfrac{1}{\sqrt 3}\Bigg) = \dfrac{2\sqrt 3}{3}\Big(\dfrac{\pi}{3}-\dfrac{\pi}{6}\Big) =[/tex]
[tex]=\dfrac{2\sqrt 3}{3}\Big(\dfrac{\pi}{3}-\dfrac{\pi}{6}\Big) = \dfrac{\sqrt 3}{3}\Big(\dfrac{2\pi}{3}-\dfrac{\pi}{3}\Big) =\boxed{\dfrac{\pi\sqrt 3}{9}}[/tex]
