Definite Integral $\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$ $$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$
This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?

Note
$$\int_0^\infty e^{-xt}\cos(at)dt=\frac{x}{x^2+a^2}$$
and hence
$$\int_0^\infty e^{-xt}(\cos(at)-\cos(bt))dt=-\frac1{a^2-b^2}\frac{x}{(x^2+a^2)(x^2+b^2)}.$$
Thus
\begin{eqnarray}
\int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx&=&2\int_{0}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx\\
&=&-\frac2{a^2-b^2}\int_{0}^\infty\left(\int_{0}^\infty e^{-xt}(\cos(at)-\cos(bt))dt\right)\sin xdx\\
&=&-\frac1{a^2-b^2}\int_{0}^\infty(\cos(at)-\cos(bt))\left(\int_{0}^\infty e^{-xt}\sin x dx\right)dt\\
&=&-\frac2{a^2-b^2}\int_{0}^\infty\frac{\cos(at)-\cos(bt)}{t^2+1}dt\\
&=&-\frac2{a^2-b^2}\left(\frac{\pi}{2e^a}-\frac{\pi}{2e^b}\right)\\
&=&-\frac1{a^2-b^2}\left(\frac{\pi}{e^a}-\frac{\pi}{e^b}\right).
\end{eqnarray}
Here we used the following result
$$ \int_0^\infty\frac{\cos(a t)}{t^2+1}dt=\frac{\pi}{2e^a}. $$
