Consider the path $\mu:[0,2\pi]\rightarrow\mathbb{C}, t \mapsto \left\{\begin{matrix} Re^{it} & \text{if } 0 \leq t\leq \pi\\ \frac{R}{\pi}(2t-3\pi) & \text{if } \pi \leq t\leq 2\pi\\ \end{matrix}\right. $
(It corresponds to the path with image $[-R,+R]\cup\left \{ \text{upper-half of the circle of radius R} \right \}$
Now, the function $f(z)=\frac{1}{z^2+z+1}$. I have to prove that when R goes to $\infty$, the integral $\oint_{\mu}f(z)dz$ vanishes to 0.
I have tried to utilize the Estimation Lemma. So, $\left| \oint_{\mu}f(z)dz\right|\leq \text{length}(\mu)\cdot\max_{z\in D_R}\left|f(z)\right|$, but I can't find the max.
ERROR: I meant the integral over the arc vanishes not the part on the real-axis
The integral along the semi-circular part of the contour, $Re^{i[0,\pi]}$ vanishes, but the integral along the $[-R,R]$ part does not vanish. $$ \int_\mu\frac1{z^2+z+1}\,\mathrm{d}z =\int_{-R}^R\frac1{x^2+x+1}\,\mathrm{d}x+\int_{Re^{i[0,\pi]}}\frac1{z^2+z+1}\,\mathrm{d}z\\ $$ We can use your estimate that the length of the arc is $\pi R$ and the triangle inequality gives us that $\left|z^2+z+1\right|\ge\left|z^2\right|-\left|z+1\right|\ge R^2-R-1$. Therefore, $$ \left|\,\int_{Re^{i[0,\pi]}}\frac1{z^2+z+1}\,\mathrm{d}z\,\right|\le\frac{\pi R}{R^2-R-1} $$ As $R\to\infty$, $\frac{\pi R}{R^2-R-1}\to0$.
We can then use the Residue Theorem to compute the integral along $\mu$ and thus get the integral along the real line $$ \int_{-\infty}^\infty\frac1{x^2+x+1}\,\mathrm{d}x=2\pi i\operatorname*{Res}_{z=\frac{-1+i\sqrt3}2}\left(\frac1{z^2+z+1}\right) $$