= Solution
For a positively oriented simple closed <contour>, the <residue theorem> states that the integral of a function holomorphic on and inside the <contour> except at finitely many isolated interior singularities is $2\pi i$ times the sum of its <residues>. There must be no singularity on the <contour>.
Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two <simple poles>
$$
a=\frac12+\frac{i\sqrt3}{2},\qquad b=-\frac12+\frac{i\sqrt3}{2}.
$$
The other two <poles> are in the lower half-plane. Since $D'(z)=4z^3+2z=2z(2z^2+1)$,
$$
\operatorname{Res}_{a}g=\frac{e^{ia}}{2i\sqrt3\,a},\qquad
\operatorname{Res}_{b}g=-\frac{e^{ib}}{2i\sqrt3\,b}.
$$
Writing $h=\sqrt3/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies to
$$
\frac{e^{ia}}a-\frac{e^{ib}}b
=e^{-h}\left[\left(\frac12-ih\right)e^{i/2}+\left(\frac12+ih\right)e^{-i/2}\right]
=e^{-h}\left[\cos\frac12+\sqrt3\sin\frac12\right].
$$
Thus for every $R>1$ the closed-contour integral is
$$
\oint_{L_R\cup C_R}g(z)\,dz
=\frac\pi{\sqrt3}e^{-\sqrt3/2}\left[\cos\frac12+\sqrt3\sin\frac12\right].
$$
To remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\operatorname{Im}z}\leq1$ and $|z^4+z^2+1|\geq R^4-R^2-1$. For sufficiently large $R$, the <complex line integral estimate> gives
$$
\left|\int_{C_R}g(z)\,dz\right|
\leq\frac{\pi R}{R^4-R^2-1}\longrightarrow0.
$$
The real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore
$$
\boxed{\int_{-\infty}^{\infty}\frac{\cos x}{x^4+x^2+1}\,dx
=\frac\pi{\sqrt3}e^{-\sqrt3/2}\left[\cos\frac12+\sqrt3\sin\frac12\right]}.
$$
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