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 _______.

The integrand is $$f(x)=\left[\sqrt{\frac{10x}{x+1}}\right]$$ where $$[\,\cdot\,]$$ denotes the greatest-integer (floor) function.

First examine the real-valued part inside the floor:
$$g(x)=\sqrt{\frac{10x}{x+1}},\qquad 0\le x\le 10$$

Because
$$g(x)^2=\frac{10x}{x+1}$$
we have
$$\frac{d}{dx}\bigl(g(x)^2\bigr)=\frac{10(x+1)-10x}{(x+1)^2}=\frac{10}{(x+1)^2}\gt0,$$
so $$g(x)$$ is strictly increasing on the entire interval. Hence $$f(x)$$ will step up only once each time $$g(x)$$ crosses an integer.

Find the points where $$g(x)=k$$ for integers $$k=1,2,3$$ (the maximum value of $$g(x)$$ at $$x=10$$ is $$\frac{10}{\sqrt{11}}\approx3.015\lt 4$$, so $$k=0,1,2,3$$ are the only relevant integers).

Require $$g(x)\lt 1$$.
$$\frac{10x}{x+1}\lt 1\;\Longrightarrow\;10x\lt x+1\;\Longrightarrow\;9x\lt 1\;\Longrightarrow\;x\lt \frac19.$$
So $$f(x)=0$$ on $$[0,\frac19)$$.

Require $$1\le g(x)\lt 2$$.
Lower bound gives $$x\ge\frac19$$.
Upper bound: $$\frac{10x}{x+1}\lt 4\;\Longrightarrow\;10x\lt 4x+4\;\Longrightarrow\;6x\lt 4\;\Longrightarrow\;x\lt \frac23.$$
Therefore $$f(x)=1$$ on $$\bigl[\frac19,\frac23\bigr).$$

Require $$2\le g(x)\lt 3$$.
Lower bound gives $$x\ge\frac23$$.
Upper bound: $$\frac{10x}{x+1}\lt 9\;\Longrightarrow\;10x\lt 9x+9\;\Longrightarrow\;x\lt 9.$$
Thus $$f(x)=2$$ on $$\bigl[\frac23,9\bigr).$$

Require $$3\le g(x)\lt 4$$.
The lower bound gives $$x\ge9$$. The upper inequality $$\frac{10x}{x+1}\lt 16$$ is automatically satisfied for all positive $$x$$, so within our domain we only need $$x\le10$$.
Hence $$f(x)=3$$ on $$[9,10].$$

Now integrate over each sub-interval:

$$I=\int_{0}^{1/9}0\,dx+\int_{1/9}^{2/3}1\,dx+\int_{2/3}^{9}2\,dx+\int_{9}^{10}3\,dx$$
$$I=0+\Bigl(\frac23-\frac19\Bigr)\cdot1+\Bigl(9-\frac23\Bigr)\cdot2+(10-9)\cdot3$$

Compute each length:
$$\frac23-\frac19=\frac{6-1}{9}=\frac59,$$
$$9-\frac23=\frac{27-2}{3}=\frac{25}{3},$$
$$10-9=1.$$

Therefore
$$I=\frac59+2\cdot\frac{25}{3}+3\cdot1=\frac59+\frac{50}{3}+3.$$

Convert to a common denominator (9):
$$I=\frac{5}{9}+\frac{150}{9}+\frac{27}{9}=\frac{182}{9}.$$

Finally,
$$9I=9\cdot\frac{182}{9}=182.$$

Hence the required value of $$9I$$ is 182.