How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea: 
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because 
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
converges
then I can't,Thank you 

Just like the comments given by @heropup and @Random Variable, this problem can be solved by using residue theorem and Jordan's lemma.
Note that the denominator of the integrand can be factorize by 
\begin{equation}
z^4+z^2+1=(z^2-z+1)(z^2+z+1)
\end{equation}
It have four poles that 
\begin{equation}
z_{1,2}=\pm\frac{1}{2}+\frac{\sqrt{3}}{2}i\\
z_{3,4}=\pm\frac{1}{2}-\frac{\sqrt{3}}{2}i\\
\end{equation}
Only $z_1$ and $z_2$ are in upper half plane. Then, from residue theorem and Jordan's lemma, we have:
\begin{equation}
I=\int_{-\infty}^{+\infty} \frac{x^3\sin(x)}{x^4+x^2+1}dx\\
=\Im(2\pi i\sum_{k=1,2}\mathrm{Res}_{z=z_k}\frac{z^3e^{iz}}{z^4+z^2+1})\\
=-\frac{1}{3}e^{-\sqrt{3}/2}\sin(\frac{1}{2})\pi\sqrt{3}+e^{-\sqrt{3}/2}\cos(\frac{1}{2})\pi
\end{equation}
The approximation of the result is $0.7938888330$. You can check it by some numerical method.
