11. Find the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$.
11. The domain of the function is $\{x \mid x \geqslant 1$ or $x \leqslant-1\}$.
(1) It is easy to see that the function $y=x^{2}+x \sqrt{x^{2}-1}$ is an increasing function on $[1,+\infty)$. Therefore, when $x \geqslant 1$, $y \geqslant 1$.
(2) When $x \leqslant-1$,
$$
y=x\left(x+\sqrt{x^{2}-1}\right)=\frac{x}{x-\sqrt{x^{2}-1}}=\frac{1}{1+\sqrt{1-\frac{1}{x^{2}}}} .
$$

Given $x \leqslant-1$, we know
$$
0 \leqslant 1-\frac{1}{x^{2}} < \frac{1}{2}.
$$
Thus, the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$ is $\left(\frac{1}{2},+\infty\right)$.