1. We need to evaluate $I=9\int_0^9 \left[\sqrt{\frac{10x}{x+1}}\right]dx.$ Let $f(x)=\sqrt{\frac{10x}{x+1}}.$ Since $[\,\cdot\,]$ is the greatest integer function, we first find where $f(x)$ crosses integers.

2. Note that for $x\in[0,9]$, $\frac{10x}{x+1}<10 \quad\Rightarrow\quad f(x)<\sqrt{10}<4.$ Also $f(0)=0$ and $f(x)$ is increasing on $[0,9]$ because $\dfrac{10x}{x+1}$ is increasing.

So $[f(x)]$ can only take values $0,1,2,3$.

3. Find the intervals where these values occur by solving $n\le f(x)<n+1.$ Equivalently, solve the thresholds $f(x)=1,2,3$.

Since $f(x)^2=\frac{10x}{x+1},$ we solve:

• For $f(x)=1$: $\frac{10x}{x+1}=1\implies 10x=x+1\implies 9x=1\implies x=\frac19.$

• For $f(x)=2$: $\frac{10x}{x+1}=4\implies 10x=4x+4\implies 6x=4\implies x=\frac23.$

• For $f(x)=3$: $\frac{10x}{x+1}=9\implies 10x=9x+9\implies x=9.$

4. Therefore,

• $[f(x)]=0$ for $0\le x<\frac19$,
• $[f(x)]=1$ for $\frac19\le x<\frac23$,
• $[f(x)]=2$ for $\frac23\le x<9$,
• At $x=9$, $f(9)=3$, but a single point does not affect the integral.

So $\int_0^9 [f(x)]dx=\int_0^{1/9}0\,dx+\int_{1/9}^{2/3}1\,dx+\int_{2/3}^{9}2\,dx.$

5. Compute: $\int_{1/9}^{2/3}1\,dx=\frac23-\frac19=\frac{6-1}{9}=\frac59,$ $\int_{2/3}^{9}2\,dx=2\left(9-\frac23\right)=2\cdot\frac{25}{3}=\frac{50}{3}.$ Hence $\int_0^9 [f(x)]dx=\frac59+\frac{50}{3}=\frac59+\frac{150}{9}=\frac{155}{9}.$

Therefore, $I=9\cdot \frac{155}{9}=155.$

6. Final answer: $\boxed{155}$

Comparison with stored correct answer: the derived answer is $155$, which matches the stored correct answer.