= Solution
For a <test function> $f\in\mathcal D(\mathbb R)$, define the proposed integral by reversing the order of integration:
$$
\langle I,f\rangle
=\int_{\mathbb R}
\left(\int_{\mathbb R}f(x)e^{ix\sqrt{\theta^2+1}}\,dx\right)d\theta
=\int_{\mathbb R}
\widehat f\bigl(-\sqrt{\theta^2+1}\bigr)\,d\theta.
$$
The <Fourier transform> of a test function is a <Schwartz function>, while $\sqrt{\theta^2+1}\asymp1+|\theta|$. The final integral is consequently <absolute convergence>[absolutely convergent]. Repeated <integration by parts> in $x$ bounds it by finitely many <seminorms> of the test function, so it defines a <continuous dual space>[continuous linear functional] on $\mathcal D(\mathbb R)$.
Equivalently, put $\lambda=\sqrt{\theta^2+1}$ on the two half-lines. Then
$$
\langle I,f\rangle
=2\int_1^\infty
\widehat f(-\lambda)
\frac{\lambda}{\sqrt{\lambda^2-1}}\,d\lambda.
$$
The density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular <tempered distribution>. Its inverse <Fourier transform> is exactly the proposed oscillatory integral. Thus
$$
\boxed{\int_{\mathbb R}e^{ix\sqrt{\theta^2+1}}\,d\theta
\in\mathcal D'(\mathbb R)}.
$$
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