3. Find the range of the function $y=\frac{2+x}{1+\sqrt{1-x^{2}}}+\frac{1-\sqrt{1-x^{2}}}{x}$.
3. From the problem, we know that $-1 \leqslant x \leqslant 1$, and $x \neq 0$. Let $x=\sin \theta, \theta \in\left[-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]$, then $y=\frac{2(1+x)}{1+\sqrt{1-x^{2}}}=\frac{2(1+\sin \theta)}{1+\cos \theta}$. Therefore, $\frac{y}{2}=\frac{1+\sin \theta}{1+\cos \theta}$, and $\theta \neq 0$.

In the Cartesian coordinate system, by the slope formula, $\frac{y}{2}$ is the slope of the line connecting the points $(-1,-1)$ and $(\cos \theta, \sin \theta)$, so, using geometric properties, we get $\frac{y}{2} \in\left[0, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right]$, which means $y \in[0,1) \cup(1, 4]$.