1. We need to evaluate $I=\int_0^{10}\left[\sqrt{\frac{10x}{x+1}}\right]dx.$ Let $f(x)=\sqrt{\frac{10x}{x+1}}.$ Since $x\in[0,10]$, the quantity $\frac{10x}{x+1}$ increases with $x$, so $f(x)$ is increasing.

2. Find the range of $f(x)$ on $[0,10]$: $f(0)=0,\qquad f(10)=\sqrt{\frac{100}{11}}<\sqrt{9}=3.$ So $0\le f(x)<3.$ Hence $\left[f(x)\right]$ can take only the values $0,1,2$.

3. We now find where $f(x)$ crosses $1$ and $2$.

For $f(x)\ge 1$: $\sqrt{\frac{10x}{x+1}}\ge 1$ $\frac{10x}{x+1}\ge 1$ $10x\ge x+1$ $9x\ge 1$ $x\ge \frac19.$ So:

• $[f(x)]=0$ for $0\le x<\frac19$.

For $f(x)\ge 2$: $\sqrt{\frac{10x}{x+1}}\ge 2$ $\frac{10x}{x+1}\ge 4$ $10x\ge 4x+4$ $6x\ge 4$ $x\ge \frac23.$ So:

• $[f(x)]=1$ for $\frac19\le x<\frac23$,
• $[f(x)]=2$ for $\frac23\le x\le 10$.

4. Therefore, $I=\int_0^{1/9}0\,dx+\int_{1/9}^{2/3}1\,dx+\int_{2/3}^{10}2\,dx.$ Compute each part: $\int_{1/9}^{2/3}1\,dx=\frac23-\frac19=\frac{6-1}{9}=\frac59,$ $\int_{2/3}^{10}2\,dx=2\left(10-\frac23\right)=2\cdot\frac{28}{3}=\frac{56}{3}.$ Hence $I=\frac59+\frac{56}{3}=\frac59+\frac{168}{9}=\frac{173}{9}.$

5. So, $9I=173.$

6. Comparison with stored answer: The stored correct answer is $182$, but our computed value is $173$. Thus, the stored answer appears to be incorrect.