Consider $f(z) = e^{-ax^2}$, and let $C$ be the rectangular contour with vertices at $-R,R,R+ib$ and $-R+ib$.
As $R \to \infty$, the two sides parallel to the imaginary axis disappear, then we have, by Cauchy's theorem:
$$
0 = \oint_C f(z) \, dz = \int_{-\infty}^\infty f(x)\, dx + \int_{\infty}^{-\infty} e^{-a(x+ib)^2} \implies\\
\int_{-\infty}^\infty e^{-ax^2 -2 a i x b +a b^2}\, dx = \int_{-\infty}^\infty f(x)\, dx = \sqrt{\frac{\pi}{a}} \implies \\
\int_{-\infty}^\infty e^{-ax^2 -2 a i x b}\, dx = \sqrt{\frac{\pi}{a}}e^{-ab^2}
$$
Taking real parts of both sides, the result follows.
Note that the integral of interest can be written
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(\int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2+2x+1}dx\right) \tag 1$$
Let's examine the integral
$$\oint_C\frac{ze^{iz}}{z^2+2z+1}dz$$
where $C$ is the closed contour comprised of
$(i)$ the real-line segment from $(-R,0)$ to $(R,0)$
$(ii)$ the semicircle $C_R$ in the upper-half plane, centered at the origin with radius $R$.
Thus, we can write
$$\begin{align}
\oint_C\frac{ze^{iz}}{z^2+2z+1}dz&=\int_{-R}^{R}\frac{xe^{ix}}{x^2+2x+1}dx+
\int_{C_R}\frac{ze^{iz}}{z^2+2z+1}dz \tag 2\\\\
\end{align}$$
Note that in the limit as $R\to \infty$, the imaginary part of first integral on the right-hand side of $(2)$ equals the Cauchy Principal Value of the integral of interest in $(1)$. The second integral on the right-hand side of $(2)$ can be shown using Jordan's Lemma to go to zero in the limit as $R\to \infty$. Thus, from the residue theorem we have
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(2\pi i \,\text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)\right)\tag 3$$
The residue in $(3)$ can be evaluated as
$$\begin{align}
\text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)&=
\lim_{z\to -1+i}\frac{(z+1-i)ze^{iz}}{(z+1-i)(z+1+i)}\\\\
&= \frac{(-1+i)e^{i(-1+i)}}{2i} \\\\
&=\frac{e^{-1}}{2i}\sqrt{2}e^{i(3\pi/4-1)}\tag 4
\end{align}$$
Substituting $(4)$ into $(3)$ reveals that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\pi e^{-1}(\cos 1 -\sin 1)}$$
Best Answer
Consider following parametric integral for $\alpha \ge0$ $$I(\alpha )=\int_{-\infty}^{\infty}\frac{1-\cos \alpha x}{x^2}\,\mathrm dx$$ We have $I(0)=0$ $$I'(\alpha )=\int_{-\infty}^{\infty}\frac{\sin \alpha x}{x}\,\mathrm dx=\pi$$ $$I(\alpha )=\pi \alpha +c$$ Then $I(0)=\pi \cdot0+c=0 \implies I(a)=\pi a$ $$I(\alpha )=\int_{-\infty}^{\infty}\frac{1-\cos \alpha x}{x^2}\,\mathrm dx=\pi \alpha $$ $$I(1)=\pi$$