Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 +  \sqrt {3}i/2)$.
Let w=$(-1/2 +  \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answer is in terms of complex number. I wanna ask whether my method is correct and seek for another method. Thank you.

\begin{eqnarray}
\int_{0}^{1} \dfrac{1}{x^2+x+1} dx &=& \int_{0}^{1} \dfrac{1}{(x+1/2)^2 +3/4} dx \\
&=& \int_{1/2}^{3/2} \dfrac{1}{y^2 +3/4}dy \\
&=& \dfrac{4}{3}\int_{1/2}^{3/2} \dfrac{1}{(\dfrac{2}{\sqrt{3}}y)^2 + 1} dy \\
&=& \dfrac{\sqrt{3}}{2} \dfrac{4}{3} \int_{\sqrt{3}/3}^{\sqrt{3}} \dfrac{1}{z^2+1} dz\\
&=& 2 \dfrac{\sqrt{3}}{3} (\arctan(\sqrt{3}) - \arctan(\sqrt{3}/3))\\
&=&2 \dfrac{\sqrt{3}}{3}( \dfrac{\pi}{3} - \dfrac{\pi}{6})\\
&=& \dfrac{\sqrt{3}\pi}{9}\\
&=& 0,6045997880781
\end{eqnarray}
