Try using complex analysis!

Although contour integration could be used straight away, the fact that the function $\frac{z e^{iz}}{(z^2 + 1)^3}$ has a triple pole at $z=i$ makes the calculations involved. It is slightly simpler to intergrate by parts first, so $I ;\ = \; \int_0^\infty \frac{x \sin x}{(x^2 + 1)^3}\,dx \; = \; \Big[ - \frac{\sin x}{4(x^2 + 1)^2} \Big]_0^\infty + \tfrac14\int_0^\infty \frac{\cos x}{(x^2 + 1)^2}\,dx \; = \; \tfrac18\int_\mathbb{R} \frac{e^{ix}}{(x^2 + 1)^2}\,dx$ and so, applying contour integration to a large radius semicircle in the upper half plane, centred at the origin, we have $\begin{aligned} I & = \tfrac18 \times 2\pi i\, \mathrm{Res}_{z=i} \frac{e^{iz}}{(z^2+1)^2} \; = \; \tfrac14\pi i \left(\frac{d}{dz} \frac{e^{iz}}{(z+i)^2}\right)\Bigg|_{z=i} \\ & = \tfrac14\pi i \left(\frac{ie^{iz}}{(z+i)^2} - \frac{2e^{iz}}{(z+i)^3}\right)\Big|_{z=i} \; = \; \tfrac14\pi i \Big(-\frac{i}{4e} - \frac{i}{4e}\Big) \; = \; \frac{\pi}{8e} \end{aligned}$ making the anwswer $\boxed{0.1444659187}$ .