8 The solution set of the inequality $x^{2}+\ln x>x$ is $\qquad$ . (Express in interval notation)
$8(1,+\infty)$ Hint: Let $f(x)=x^{2}+\ln x-x$, then
$$
f^{\prime}(x)=2 x+\frac{1}{x}-1=\frac{2 x^{2}-x+1}{x}>0,
$$
$f(x)$ is an increasing function on its domain $(0,+\infty)$.
Also, $f(1)=0$, so the solution set of $f(x)>0$ is $(1,+\infty)$, which means the solution set of the inequality $x^{2}+$ $\ln x>x$ is $(1,+\infty)$.