How can I find the principal value of the following? PV∫−∞∞eixx(x2+x+1)dx

With

$\begin{array}{}p=\frac{-1+\sqrt{3}i}{2}:\\ P.V.{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x(x^2+x+1)}dx\\ ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{x^2+x+1}[\frac{1}{x+i{0}^{+}}+i\pi \delta (x)]dx\\ ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ix}}{(x+i{0}^{+})(x-p)(x-\bar{p})}dx+i\pi \\ =i\pi +2\pi i\frac{e^{ip}}{p(p-\bar{p})}=i\pi +2\pi i\frac{e^{-i/2-\sqrt{3}/2}}{[(-1+\sqrt{3}i)/2](i\sqrt{3})}\\ =i\pi +2\pi \frac{\sqrt{3}}{3}{e}^{-i/2-\sqrt{3}/2}\frac{-1-\sqrt{3}i}{2}\\ =(i-\frac{\sqrt{3}+3i}{3}{e}^{-i/2-\sqrt{3}/2})\pi \approx -1.3030+2.3477i\end{array}$

You can use the residue theorem. Consider the complex contour integral
${\oint }_{C}dz\frac{e^{iz}}{z(1+z+z^2)}$
where C is a semicircle of radius R in the upper half plane with a small semicircular detour about the origin into the upper half plane of radius $ϵ$. Then the contour integral is equal to
$PV{\int }_{-R}^{R}dx\frac{e^{ix}}{x(1+x+x^2)}+iϵ{\int }_{\pi }^{0}d\varphi e^{i\varphi }\frac{e^{iϵe^{o\varphi }}}{ϵe^{i\theta }(1+ϵe^{i\varphi }+{ϵ}^2{e}^{i2\varphi }}$
As $R\to \mathrm{\infty }$, the third integral vanishes as $1/R^3$. As $ϵ\to 0$, the second integral becomes $-i\pi$. By the residue theorem, the contour integral is equal to $i2\pi$ times the residue of any poles inside C. The poles of the integrand are at
$z^2+z+1=0\Rightarrow z=\frac{1\pm i\sqrt{3}}{2}$
Only $z_{+}$ is inside C, with residue
$\frac{e^{-i/2}{e}^{-\sqrt{3}/2}}{e^{i2\pi /3}(i\sqrt{3})}$
Thus,
$PV{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\frac{e^{ix}}{x(1+x+x^2)}=\frac{2\pi }{\sqrt{3}}{e}^{-\sqrt{3}/2}{e}^{-i(2\pi /3+1/2)}+i\pi$

$PV{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{x(1+x+x^2)}dx=PV{\int }_0^{+\mathrm{\infty }}(\frac{2i(1+x^2)}{1+x^2+x^4}\cdot \frac{\sin x}{x}-\frac{2}{1+x^2+x^4}\cos (x))$
and both integrals giving the real and imaginary part can be computed by using the Laplace transform. We have:
${\int }_0^{+\mathrm{\infty }}\frac{2\cos x}{1+x^2+x^4}dx=\frac{\pi }{3}{e}^{-\frac{\sqrt{3}}{2}}(3\sin \frac{1}{2}+\sqrt{3}\cos \frac{1}{2})$
and:
${\int }_0^{+\mathrm{\infty }}\frac{1+x^2}{1+x^2+x^4}\cdot \frac{\sin x}{x}dx=\frac{\pi }{6}(3+2\sqrt{3}{e}^{-\frac{\sqrt{3}}{2}}\sin \frac{1}{2})$