For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. If $I = \int_{0}^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx$, then the value of $9I$ is -----.

$I=\int_{0}^{10}\left[\sqrt{\dfrac{10x}{x+1}}\right]dx$

$\text{Find } I, \text{ then } 9I$

$\phi(x)=\sqrt{\dfrac{10x}{x+1}}$

$\phi(0)=0,\quad \phi(10)=\sqrt{\dfrac{100}{11}}\approx 3.01$

$\therefore [\phi(x)] \in \{0,1,2,3\}$

$\phi(x)=k \Leftrightarrow \dfrac{10x}{x+1}=k^{2}$

$\Rightarrow x=\dfrac{k^{2}}{10-k^{2}}$

$k=1:\ x=\dfrac{1}{9},\qquad k=2:\ x=\dfrac{4}{6}=\dfrac{2}{3}$

$k=3:\ x=\dfrac{9}{1}=9,\qquad k=4:\ x=\dfrac{16}{-6}<0\ \text{(invalid)}$

$[0,\tfrac{1}{9}): 0;\quad [\tfrac{1}{9},\tfrac{2}{3}): 1$

$[\tfrac{2}{3},9): 2;\quad [9,10]: 3$

$I = 0\cdot\tfrac{1}{9} + 1\cdot\left(\tfrac{2}{3}-\tfrac{1}{9}\right) + 2\cdot\left(9-\tfrac{2}{3}\right) + 3\cdot 1$

$= \dfrac{5}{9} + \dfrac{50}{3} + 3 = \dfrac{182}{9}$

$9I = 9\cdot\dfrac{182}{9} = 182$

$\boxed{\text{Answer: } 182}$