Let a differentiable function f satisfy f (x) + ∫ _ 3 ^ x (f...

Given:

$f\left(x\right) + {∫}_{3}^{x}\dfrac{f\left(t\right)}{t}dt = \sqrt{x + 1}$

$⇒ {f}^{'}\left(x\right) + \dfrac{f\left(x\right)}{x} = \dfrac{1}{2\sqrt{x + 1}}$

Put $y = f\left(x\right)$, then

$\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{1}{2\sqrt{x + 1}}$

So, $I . F . = {e}^{∫ \dfrac{dx}{x}} = {e}^{\ln \left|x\right|} = x$

Hence, solution is

$xy = \dfrac{1}{2} ∫ \dfrac{x}{\sqrt{x + 1}}dx$

$⇒ xy = \dfrac{1}{2} ∫ \left(\dfrac{x + 1 - 1}{\sqrt{x + 1}}\right)dx$

$⇒ xy = \dfrac{1}{2} ∫ \left(\sqrt{x + 1} - \dfrac{1}{\sqrt{x + 1}}\right)dx$

$⇒ xy = \dfrac{1}{2}\left(\dfrac{2}{3}\left(x + 1\right)\dfrac{3}{2} - 2\sqrt{x + 1}\right) + C$

$⇒ xy = \dfrac{1}{3}\left(x + 1\right)\dfrac{3}{2} - \sqrt{x + 1} + C$

Put $x = 3$, then $f\left(3\right) = 2$

So,

$6 = \dfrac{8}{3} - 2 + C ⇒ C = \dfrac{16}{3}$

Hence,

$xy = \dfrac{1}{3}\left(x + 1\right)\dfrac{3}{2} - \sqrt{x + 1} + \dfrac{16}{3}$

So, put $x = 8$, then

$8f\left(8\right) = \dfrac{27}{3} - 3 + \dfrac{16}{3}$

$⇒ f\left(8\right) = \dfrac{34}{3 \times 8}$

$⇒ 12f\left(8\right) = 17$