Let a differentiable function $f$ satisfy $f\left(x\right)+{\int }_3^x \frac{f\left(t\right)}{t}dt=\sqrt{x+1},x\ge 3$. Then 12 $f\left(8\right)$ is equal to:

$f\left(x\right)+{\int }_3^x \frac{f\left(t\right)}{t}dt=\sqrt{x+1};x\ge 3\dots .\left(1\right)$ differentiating with respect to x
$f^{\mathrm{\prime }}\left(x\right)+\frac{f\left(x\right)}{x}=\frac{1}{2\sqrt{x+1}}$   (linear form)
Integrating factor is $e^{\int \frac{1}{x}dx}=x$
$\therefore$ General solution is $xf\left(x\right)=\int \frac{x}{2\sqrt{x+1}}dx\Rightarrow x\mathrm{f}\left(x\right)=\frac{(x+1{)}^{3/2}}{3}-\sqrt{x+1}+C$
Put  x =3 in equation (1) we have $f\left(3\right)=2$
$\therefore 3\left(2\right)=\frac{1}{3}\left(8\right)-2+C\Rightarrow C=\frac{16}{3}$
$$\begin{gathered} xf\left(x\right)=\frac{(x+1{)}^{3/2}}{3}-\sqrt{x+1}+\frac{16}{3} \\ if x=8 \\ \Rightarrow 8f\left(8\right)=9-3+\frac{16}{3} \\ \Rightarrow 4f\left(8\right)=3+\frac{8}{3} \\ \Rightarrow 12f\left(8\right)=17 \end{gathered}$$