Completing the square before integrating

Completing the square. $x^2+x+1=\left(x+\frac12\right)^2+\frac34=\frac34\left[\left(\frac{2x+1}{\sqrt3}\right)^2+1\right]$.

Antiderivative. With $u=\frac{2x+1}{\sqrt3}$, $du=\frac2{\sqrt3}dx$:
$$\int\frac{dx}{x^2+x+1}=\frac43\cdot\frac{\sqrt3}2\int\frac{du}{u^2+1}=\frac2{\sqrt3}\arctan\frac{2x+1}{\sqrt3}+C.$$

Definite integral.
$$\frac2{\sqrt3}\left(\arctan\sqrt3-\arctan\frac1{\sqrt3}\right)=\frac2{\sqrt3}\left(\frac\pi3-\frac\pi6\right)=\frac\pi{3\sqrt3}\approx0.6046.$$