111 The increasing interval of the function $y=\frac{x-\sqrt{3}}{\sqrt{1+x^{2}}}$ is $\qquad$
$111\left(-\frac{\sqrt{3}}{3},+\infty\right)$. Let $x=\tan \theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then
$$
y=\frac{x-\sqrt{3}}{\sqrt{1+x^{2}}}=2 \sin \left(\theta-\frac{\pi}{3}\right) .
$$

It is easy to see that when $x \in\left(-\frac{\sqrt{3}}{3},+\infty\right)$, the function is monotonically increasing.