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 

Note that
$$I(a) = \int_{-\infty}^\infty \frac{t\sin at} {t^2+1}dt
= \pi e^{-a}
$$
(derived from $I’’(a)=I(a)$, $I(0)= -I’(0) =\pi$).  Then
\begin{align}
\int_{-\infty}^{\infty}\frac{x^3\sin x}{x^4+x^2+1}\> dx
= &\frac12\int_{-\infty}^{\infty}
 \overset{x=\frac{\sqrt3t -1}2}{\frac{(x+1)\sin x}{x^2+x+1}} +\overset{x= \frac{\sqrt3t +1}2}{\frac{(x-1)\sin x}{x^2-x+1}} d x\\
=&\cos\frac1{2} \int_{-\infty}^{\infty} \frac{t\sin \frac {\sqrt3}{2}t}{t^2+1} dt - \frac1{\sqrt3}\sin\frac1{2} \int_{-\infty}^{\infty} \frac{\cos\frac {\sqrt3}{2}t}{t^2+1} dt \\
=& \cos\frac1{2}  I(\frac {\sqrt3}{2})
+\frac1{\sqrt3}\sin\frac1{2} I’(\frac {\sqrt3}{2})\\
=&\pi \>e^{-\frac {\sqrt3}{2} }\left( \cos\frac12-\frac1{\sqrt3} \sin\frac12\right)\\
\end{align}
