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 
 A: 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.
A: Follow the steps :
1) use partial fractions.
2) use Trigonometric integrals.
A: May I confess that I sawer nicer integrals ?  
Anyway, being patient and using Mhenni Benghorbal's suggestions plus a series of integrations by parts as well as a few changes of variables, the antiderivative is
$$\frac{1}{24} e^{-\sqrt[6]{-1}} \left(e^{\sqrt{3}} \left(\left(\sqrt{3}-3 i\right)
   \left(\text{Ei}\left(i x+(-1)^{5/6}\right)-\text{Ei}\left(\frac{1}{2} \left(-2 i
   x-\sqrt{3}+i\right)\right)\right)+\left(\sqrt{3}+3 i\right) e^i
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x-\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x-\sqrt{3}-i\right)\right)\right)\right)-\left(\sqrt{3}-3 i\right) e^i
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x+\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x+\sqrt{3}-i\right)\right)\right)+\left(\sqrt{3}+3 i\right)
   \left(\text{Ei}\left(\frac{1}{2} \left(-2 i
   x+\sqrt{3}+i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
   x+\sqrt{3}+i\right)\right)\right)\right)$$ Once integrated between $-\infty$ and $\infty$, again after some patience, I arrived to $$\frac{\left(-i+\sqrt{3}+\left(\sqrt{3}+i\right) e^i\right) e^{-\sqrt[6]{-1}} \pi }{2
   \sqrt{3}}$$ that is to say $$-\frac{1}{3} e^{-\frac{\sqrt{3}}{2}} \pi  \left(\sqrt{3} \sin
   \left(\frac{1}{2}\right)-3 \cos \left(\frac{1}{2}\right)\right)$$ which reduces to $$\frac{2}{\sqrt{3}} e^{-\sqrt{3}/2} \pi \cos \frac{\pi+3}{6}$$ which corresponds to heropup's result.   
Thanks to Mhenni Benghorbal and heropup for their suggestions and ideas.
A: 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}
