= Solution
<Jordan lemma> states that if $a>0$ and $g$ is analytic in the upper half-plane outside a fixed circle, with
$$
M_R=\max_{0\leq\theta\leq\pi}|g(Re^{i\theta})|\longrightarrow0,
$$
then on the upper semicircle $C_R$,
$$
\int_{C_R}e^{iaz}g(z)\,dz\longrightarrow0.
$$
Indeed, $|e^{iaRe^{i\theta}}|=e^{-aR\sin\theta}$. Using symmetry and $\sin\theta\geq2\theta/\pi$ on $0\leq\theta\leq\pi/2$ gives
$$
\left|\int_{C_R}e^{iaz}g(z)\,dz\right|
\leq2RM_R\int_0^{\pi/2}e^{-2aR\theta/\pi}\,d\theta
\leq\frac{\pi M_R}{a}\longrightarrow0.
$$
The <residue> of $f$ at an isolated singularity $a$ is the coefficient of $(z-a)^{-1}$ in its <Laurent series>, equivalently
$$
\operatorname{Res}(f,a)=\frac1{2\pi i}\oint f(z)\,dz.
$$
If $f(z)=g(z)/(z-a)^k$, the Taylor expansion of the analytic function $g$ shows that
$$
\boxed{\operatorname{Res}(f,a)=\frac{g^{(k-1)}(a)}{(k-1)!}}.
$$
For the real integral, apply <contour integration> in the upper half-plane to
$$
F(z)=\frac{z^3e^{iz}}{(1+z^2)^2}.
$$
Jordan's lemma removes the semicircular arc. The only enclosed singularity is the double pole at $z=i$, whose residue is
$$
\operatorname{Res}(F,i)
=\left.\frac d{dz}\left(\frac{z^3e^{iz}}{(z+i)^2}\right)\right|_{z=i}
=\frac1{4e}.
$$
The <residue theorem> therefore gives
$$
\int_{-\infty}^{\infty}\frac{x^3e^{ix}}{(1+x^2)^2}\,dx
=\frac{\pi i}{2e}.
$$
Its real part vanishes because $x^3\cos x/(1+x^2)^2$ is an <odd function>, while its imaginary part is the requested integral. Hence
$$
\boxed{\int_{-\infty}^{\infty}\frac{x^3\sin x}{(1+x^2)^2}\,dx=\frac\pi{2e}}.
$$
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