Example 1 Find the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$. ${ }^{[1]}$ (2013, Hubei Provincial Preliminary of the National High School Mathematics League)
It is known that the domain of the function is $\{x \mid x \geqslant 1$ or $x \leqslant-1\}$.
Let $x=\sec \theta\left(0 \leqslant \theta \leqslant \pi, \theta \neq \frac{\pi}{2}\right)$. Then $y=\sec ^{2} \theta+\sec \theta \cdot \sqrt{\sec ^{2} \theta-1}=\frac{1}{\cos ^{2} \theta}+\frac{|\tan \theta|}{\cos \theta}$.
(1) When $0 \leqslant \theta<\frac{\pi}{2}$, $0 \leqslant \sin \theta<1$.

Thus, $y=\frac{1}{\cos ^{2} \theta}+\frac{|\tan \theta|}{\cos \theta}=\frac{1+\sin \theta}{\cos ^{2} \theta}=\frac{1}{1-\sin \theta} \geqslant 1$.
When $\theta=0$, $y=1$;
When $\theta \rightarrow \frac{\pi}{2}$, $y \rightarrow+\infty$.
(2) When $\frac{\pi}{2}<\theta \leqslant \pi$, $0 \leqslant \sin \theta<1$.

Thus, $y=\frac{1}{\cos ^{2} \theta}+\frac{|\tan \theta|}{\cos \theta}=\frac{1-\sin \theta}{\cos ^{2} \theta}$
$=\frac{1}{1+\sin \theta} \in\left(\frac{1}{2}, 1\right]$.
When $\theta \rightarrow \frac{\pi}{2}$, $y \rightarrow \frac{1}{2}$;
When $\theta=\pi$, $y=1$.
In summary, the range of $y=x^{2}+x \sqrt{x^{2}-1}$ is $\left(\frac{1}{2},+\infty\right)$.