4. Let the real number $x$ be not equal to 0 and 1. Then the range of $y=$ $\frac{x^{4}-x^{2}+1}{x^{3}-x}$ is $\qquad$ .
4. $|y| \geqslant 2$.

Then $y=\frac{\left(x^{2}-1\right)^{2}+x^{2}}{x\left(x^{2}-1\right)}=\frac{x^{2}-1}{x}+\frac{x}{x^{2}-1}$, let $t_{1}=$ $\frac{x^{2}-1}{x}$, then $x^{2}-t_{1} x-1=0$. Its discriminant $\Delta_{1}=t_{1}^{2}+4>0$, which means $t_{1}$ can take any real number.

Let $t_{1}+\frac{1}{t_{1}}=t_{2}\left(t_{1} \neq 1\right)$, then $t_{1}^{2}-t_{2} t_{1}+1=0$. Let its discriminant $\Delta_{2}=t_{2}^{2}-4 \geqslant 0$, we get $\left|t_{2}\right| \geqslant 2$, i.e., $|y| \geqslant 2$. This is the range of values for $y=\frac{x^{4}-x^{2}+1}{x^{3}-x}(x \neq 0,1)$.