Fleshing out Ragib's idea: we have the substitution $\,t\to -u+\pi i\Longrightarrow dt=-du,$ , so we put
$$f(t):=\frac{t^2e^t}{1+e^{2t}}\Longrightarrow$$
$$\Longrightarrow\int_{R+\pi i}^{-R+\pi i}f(t)\,dt=-\int_{-R}^R\frac{(u^2-2\pi iu-\pi^2)e^{-u+\pi i}}{1+e^{-2u+2\pi i}}\,du=$$
$$=\int_{-R}^R\frac{u^2e^{-u}}{1+e^{-2u}}du-2\pi i\int_{-R}^R\frac{ue^{-u}}{1+e^{-2u}}\,du-\pi^2\int_{-R}^R\frac{e^{-u}}{1+e^{-2u}}\,du$$
The first integral above is just like the one on the real axis (BTW, note the signs in front of the two last integrals, which come from the factor $\,e^{\pi i}=-1\,$).
The last integral is
$$\pi^2\int_{-R}^R\frac{d(e^{-u})}{1+e^{-2u}}=\left.\pi^2\arctan e^{-u}\right|_{-R}^R=$$
$$=\pi^2\left(\arctan e^{-R}-\arctan e^R\right)\xrightarrow [R\to\infty]{}\pi^2\left(0-\frac{\pi}{2}\right)=-\frac{\pi^3}{2}$$
The second integral above is zero as the integrand is just $\,\frac{\pi iu}{\cosh u}\,$ , which is an odd function...
Denoting our integral by $\, I \,$, we thus have that
$$2I-\frac{\pi ^3}{2}=-\frac{\pi^3}{4}\Longrightarrow I=\frac{\pi^3}{8}$$
Write $\sin{x} = (e^{i x}-e^{-i x})/(2 i)$. Then consider the integral
$$PV \oint_{C_{\pm}} dx \frac{e^{\pm i z}}{z (z^2-2 z+2)} $$
where $C_{\pm}$ is a semicircular contour of radius $R$ in the upper/lower half plane with a semicircular detour into the upper/lower half plane of radius $\epsilon$. For $C_{+}$, we have
$$PV \oint_{C_{+}} dz \frac{e^{i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta} - 2 R e^{i \theta}+2)} $$
For $C_-$, we have
$$PV \oint_{C_{-}} dz \frac{e^{-i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{-i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{-\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{-i x}}{x (x^2-2 x+2)}- i R \int_0^{\pi} d\theta \, e^{-i \theta} \frac{e^{-i R e^{-i \theta}}}{R e^{-i \theta} (R^2 e^{-i 2 \theta} - 2 R e^{-i \theta}+2)} $$
In both cases, we take the limits as $R \to \infty$ and $\epsilon \to 0$. Note that, in both cases, the respective fourth integrals have a magnitude bounded by
$$\frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R^3}$$
The respective second integrals of $C_{\pm}$, on the other hand, become equal to $\mp i \frac{\pi}{2} $. Thus,
$$PV \oint_{C_{\pm}} dz \frac{e^{\pm i z}}{z (z^2-2 z+2)} = PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2}$$
On the other hand, the respective contour integrals are each equal to $\pm i 2 \pi$ times the sum of the residues of the poles inside their contours. (For $C_-$, there is a negative sign because the contour was traversed in a clockwise direction.) The poles of the denominator are at $z_{\pm}=1 \pm i$. Thus,
$$PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2} = \pm i 2 \pi \frac{e^{\pm i (1 \pm i)}}{(1 \pm i) (2) (\pm i)} $$
Taking the difference between the two results and dividing by $2 i$, we get that
$$\int_{-\infty}^{\infty} dx \frac{\sin{x}}{x (x^2-2 x+2)} = \frac{\pi}{2} \left (1+\frac{\sin{1}-\cos{1}}{e} \right ) $$
Note that we may drop the $PV$ because the difference between the integrals removes the pole at the origin.
Best Answer
With substitution $x=\tan(t/2)$ $$\int_{-\infty}^\infty \frac{e^{\frac{-1}{1+x^2}}}{1+x^2}dx=\dfrac{1}{2\sqrt{e}}\int_{-\pi}^{\pi}e^{-\frac12\cos t}\ dt= \dfrac{\pi}{\sqrt{e}}I_0(\frac12)$$