Complex Analysis Approach
Consider the function
$$f(z) = \frac{z^{\alpha}}{(z^2+1)^2}$$
Note that $z^{\alpha} = e^{\alpha \log(z)}$ and choose the branch cut on the imaginary axis where $$\log(z) = \log|z|+ i\theta , \,\,\,\theta \in (-3\pi/2,3\pi/2)$$
Consider the following contour
$$\int^{-r}_{-R}f(z) \,dz +\int^{R}_{r}f(z) \,dz+ \int_{C_R}f(z)\,dz+\int_{c_r}f(z)\,dz =2\pi i \mathrm{Res}(f,i)$$
As you showed the integrals on the semi-circles go to zero as $r \to 0, R \to \infty$ hence
$$\int^{0}_{-\infty}\frac{e^{\alpha(\log|x|+i\pi)}}{(x^2+1)^2}\,dx+\int_{0}^{\infty}\frac{e^{\alpha(\log|x|)}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$
$$(1+e^{\alpha \pi i})\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx = 2\pi i \mathrm{Res}(f,i)$$
Note that
$$\mathrm{Res}(f,i) = \lim_{z \to i}\frac{d}{dz} \frac{e^{\alpha \log(z)}}{(z+i)^2} = \lim_{z \to i} \frac{\alpha z^{\alpha-1}(z+i)-2z^{\alpha}}{(z+i)^3} = \frac{2e^{i\alpha \pi/2}(\alpha-1)}{-8 i }$$
$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx=\frac{\pi e^{i\alpha \pi/2}(1-\alpha)}{2(1+e^{\alpha \pi i}) } = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right)}$$
Real Analysis Approach
$$\int_{0}^{\infty}\frac{x^{\alpha}}{(x^2+1)^2}\,dx$$
Let $x^2 = t$
\begin{align}\frac{1}{2}\int_{0}^{\infty}\frac{t^{(\alpha-1)/2}}{(t+1)^2}\,dt &= \frac{1}{2}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(2-\frac{\alpha +1}{2}\right)
\tag{1}\\ & = \frac{(1-\alpha)}{4}\Gamma \left(\frac{\alpha +1}{2} \right)\Gamma \left(1-\frac{\alpha +1}{2}\right)
\tag{2}\\ & = \frac{\pi (1-\alpha)}{4 \sin\left((1-\alpha)\pi/2 \right)}
\tag{3}\\ & = \frac{\pi (1-\alpha)}{4\cos\left( \frac{\pi}{2}\alpha\right) }\tag{4}\end{align}
(1). $$\int^\infty_0 \frac{t^{x-1}}{(1+t)^{x+y}}\,dt = B(x,y)$$
(2). $$\Gamma(x+1) = x \Gamma(x)$$
(3). $$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$$
For $$I=\int\dfrac{\cos( u)}{u+i\pi n}\,du $$
$$u+i\pi n=v \implies I=\int \frac{\cos(v-i\pi n)}v \,dv$$ Expand the cosine
$$\cos(v-i\pi n)=\cosh (\pi n) \cos (v)+i \sinh (\pi n) \sin (v)$$
$$I=\cosh (\pi n) \int \frac{\cos(v)}v \,dv+i \sinh (\pi n) \int \frac{\sin(v)}v \,dv$$
$$I=\cosh (\pi n)\,\text{Ci}(v)+i \sinh (\pi n) \,\text{Si}(v)$$
Best Answer
I will write
$$ J_{\pm} = \int_{0}^{\infty} xe^{-x}\cos\left(x\pm\frac{x^2}{\alpha}\right)\,\mathrm{d}x $$
so that your integral takes the form $\frac{1}{2}(J_{+} + J_{-})$. Then
\begin{align*} J_{\pm} &= \operatorname{Re}\left[ \int_{0}^{\infty} x\exp\left( -x + ix \pm \frac{ix^2}{\alpha}\right) \,\mathrm{d}x. \right] \end{align*}
Now write $\mathbb{H}_{\text{right}} = \{ z \in \mathbb{C} : \operatorname{Re}(z) > 0 \}$. Then for each $a \in \mathbb{H}_{\text{right}}$, the map $z \mapsto \int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x$ is analytic on $\mathbb{H}_{\text{right}}$ and continuous on $\overline{\mathbb{H}_{\text{right}}}$. Moreover, if $a, z \in (0, \infty)$, then with $b = a^2/4z$,
\begin{align*} \int_{0}^{\infty} x e^{-ax-zx^2} \, \mathrm{d}x &= \int_{0}^{\infty} x \exp\bigg( -b \left(\frac{2z x}{a}+1\right)^2 + b \bigg) \, \mathrm{d}x \\ &= \frac{be^{b}}{z} \int_{1}^{\infty} (u-1) e^{-bu^2} \, \mathrm{d}u \\ &= \frac{be^{b}}{z} \left( \int_{1}^{\infty} u e^{-bu^2} \, \mathrm{d}u - \int_{0}^{\infty} e^{-bu^2} \, \mathrm{d}u + \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \right) \\ &= \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u. \end{align*}
Then by the principle of analytic continuation, this holds for all $a \in \mathbb{H}_{\text{right}}$ and $z \in \overline{\mathbb{H}_{\text{right}}}$. Then plugging $a = 1-i$ and $z = z_{\pm} = \pm i/\alpha$, we get $b = b_{\pm} = \mp \alpha /2 \in \mathbb{R}$.
\begin{align*} J_{\pm} &= \operatorname{Re}\bigg[ \frac{1}{2z} - \frac{\sqrt{\pi} a e^{b}}{4z^{3/2}} + \frac{be^{b}}{z} \int_{0}^{1} e^{-bu^2} \, \mathrm{d}u \bigg] \\ &= -\frac{\sqrt{\pi} \, e^{b}}{4} \operatorname{Re}\bigg[ \frac{a}{z^{3/2}} \bigg], \end{align*}
By noting that $a_{+}/z_{+}^{3/2} = -\sqrt{2}\,\alpha^{3/2}$ and $a_{-}/z_{-}^{3/2} = i\sqrt{2}\,\alpha^{3/2}$, we get
$$ J_{+} = \frac{\sqrt{2\pi}}{4} \alpha^{3/2} e^{-\alpha/2}, \qquad J_{-} = 0. $$
This complete the proof.