2. The interval of increase for the function $y=\frac{x^{2}}{x-1}(x \in \mathbf{R}, x \neq 1)$ is
(A) $x \geqslant 2$
(B) $x \leqslant 0$ or $x \geqslant 2$
(C) $x \leqslant 0$
(D) $x \leqslant 1-\sqrt{2}$ or $x \geqslant \sqrt{2}$
2. (B).

Let $t=x-1(t \neq 0)$, then $y=2+\left(t+\frac{1}{t}\right)(t \neq 0)$. If $t>0$, then $y=\left(\sqrt{t}-\frac{1}{\sqrt{t}}\right)^{2}+4$, at this time, if $t \geqslant 1$, then $\sqrt{t}-\frac{1}{\sqrt{t}} \geqslant 0, y$ is an increasing function of $t$; when $0<t \leqslant 1$, then $y=\left(\frac{1}{\sqrt{t}}-\sqrt{t}\right)^{2}+4 \cdot \frac{1}{\sqrt{t}}-\sqrt{t} \geqslant 0\lceil 1)(-\mathrm{F} t$ is decreasing, at this time $y$ is a decreasing function of $t$: If $t<0$, then $y=-\left(\sqrt{-t}-\frac{1}{1-t}\right)^{2}$, when $-t \geqslant 1$, i.e., $t \leqslant-1$, $y$ is increasing; when $0<-t \leqslant 1$, $y=-\left(\frac{1}{\sqrt{-1}}-\sqrt{-1}\right)^{2}$ is decreasing.
In summary, the increasing interval of $y$ is $x \leqslant 0$ or $x \geqslant 2$.