Let a differentiable function f satisfy f x + ∫ 3 x f t t d t = x + 1 , x ≥ 3 . Then 12 f 8 is…

Given:

$f\left(x\right)+{\int }_3^x\frac{f\left(t\right)}{t}dt=\sqrt{x+1}$

$\Rightarrow f^{\text{'}}\left(x\right)+\frac{f\left(x\right)}{x}=\frac{1}{2\sqrt{x+1}}$

Put $y=f\left(x\right)$, then

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

So, $\mathrm{I}.\mathrm{F}.=e^{\int \frac{dx}{x}}=e^{\ln \left|x\right|}=x$

Hence, solution is

$xy=\frac{1}{2}\int \frac{x}{\sqrt{x+1}}dx$

$\Rightarrow xy=\frac{1}{2}\int \left(\frac{x+1-1}{\sqrt{x+1}}\right)dx$

$\Rightarrow xy=\frac{1}{2}\int \left(\sqrt{x+1}-\frac{1}{\sqrt{x+1}}\right)dx$

$\Rightarrow xy=\frac{1}{2}\left(\frac{2}{3}{\left(x+1\right)}^{\frac{3}{2}}-2\sqrt{x+1}\right)+C$

$\Rightarrow xy=\frac{1}{3}{\left(x+1\right)}^{\frac{3}{2}}-\sqrt{x+1}+C$

Put $x=3$, then $f\left(3\right)=2$

So,

$6=\frac{8}{3}-2+C\Rightarrow C=\frac{16}{3}$

Hence,

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

So, put $x=8$, then

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

$\Rightarrow f\left(8\right)=\frac{34}{3\times 8}$

$\Rightarrow 12f\left(8\right)=17$