1. Find the range of the function $y=\frac{2+x}{1+\sqrt{1-x^{2}}}+\frac{1-\sqrt{1-x^{2}}}{x}$.
1. Solution: Let $x=\sin \theta, \theta \in\left[-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]$,

then $y=\frac{2 x+x^{2}+1-\left(1-x^{2}\right)}{x\left(1+\sqrt{1-x^{2}}\right)}=\frac{2(1+x)}{1+\sqrt{1-x^{2}}}=\frac{2(1+\sin \theta)}{1+\cos \theta}$,
since $\frac{1+\sin \theta}{1+\cos \theta}=\frac{\sin \theta-(-1)}{\cos \theta-(-1)}$ represents the slope of the line connecting a point on the unit circle (the right half of the circle excluding the intersection with the x-axis) and the fixed point $(-1,-1)$. According to the graph, $\frac{1+\sin \theta}{1+\cos \theta} \in\left[0, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 2\right]$, therefore, the range of the original function is $[0,1) \cup(1,4]$.