The value 9 ∫ _ 0 ^ 9 [ (10x)/(x + 1) ]dx , where t denotes the...

Let,

$I = {∫}_{0}^{9}\left[\sqrt{\dfrac{10x}{x + 1}}dx\right]$

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

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

$⇒ 1 = \dfrac{10x}{x + 1} ⇒ x + 1 = 10x$

$⇒ x = \dfrac{1}{9} . . . \left(ii\right)$

$⇒ 4 = \dfrac{10x}{x + 1} ⇒ 4x + 4 = 10x$

$⇒ x = \dfrac{2}{3} . . . \left(iii\right)$

$⇒ 9 = \dfrac{10x}{x + 1} ⇒ 9x + 9 = 10x$

$⇒ x = 9 . . . \left(iv\right)$

Now, from the above equations using the law ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{c}f\left(x\right) d x + {∫}_{c}^{d}f\left(x\right) d x + {∫}_{d}^{b}f\left(x\right) d x$

$⇒ I = {∫}_{0}^{\dfrac{1}{9}}\left(0\right)dx + {∫}_{\dfrac{1}{9}}^{\dfrac{2}{3}}\left(1\right)dx + {∫}_{\dfrac{2}{3}}^{9}\left(2\right)dx$

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

$⇒ I = \dfrac{155}{9}$

$⇒ 9I = 155$