The value 9 ∫ 0 9 10 x x + 1 dx , where t denotes the greatest integer less than or equal to t , is _____.

Let,

$I={\int }_0^9\left[\sqrt{\frac{10x}{x+1}}dx\right]$

The square of integers are $0, 1, 4, 9, ...$

$\Rightarrow 0=\frac{10x}{x+1}\Rightarrow x=0 ...\left(i\right)$

$\Rightarrow 1=\frac{10x}{x+1}\Rightarrow x+1=10x$

$\Rightarrow x=\frac{1}{9} ...\left(ii\right)$

$\Rightarrow 4=\frac{10x}{x+1}\Rightarrow 4x+4=10x$

$\Rightarrow x=\frac{2}{3} ...\left(iii\right)$

$\Rightarrow 9=\frac{10x}{x+1}\Rightarrow 9x+9=10x$

$\Rightarrow x=9 ...\left(iv\right)$

Now, from the above equations using the law ${\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{d}f\left(x\right)dx+{\int }_d^{b}f\left(x\right)dx$

$\Rightarrow I={\int }_0^{\frac{1}{9}}\left(0\right)dx+{\int }_{\frac{1}{9}}^{\frac{2}{3}}\left(1\right)dx+{\int }_{\frac{2}{3}}^9\left(2\right)dx$

$\Rightarrow I=\frac{2}{3}-\frac{1}{9}+2\left(9-\frac{2}{3}\right)$

$\Rightarrow I=\frac{155}{9}$

$\Rightarrow 9I=155$