Let a differentiable function $f$ satisfy $f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$. Then $12 f(8)$ is equal to :

Differentiating both sides we get, <br/><br/>$$ \begin{aligned} & f^{\prime}(x)+\frac{f(x)}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{1}{2 \sqrt{x+1}} \\\\ & \Rightarrow \mathrm{IF}=x \\\\ & \Rightarrow y x=\frac{1}{2} \int \frac{x}{\sqrt{x+1}} d x+c \\\\ & \Rightarrow y x=\frac{1}{2}\left(\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-2(x+1)^{\frac{1}{2}}\right)+c \\\\ & x y=\frac{1}{3}(x+1)^{\frac{3}{2}}-(x+1)^{\frac{1}{2}}+c \\\\ & f(3)=2 \end{aligned} $$ <br/><br/>So, $x=3, y=2$ <br/><br/>$\Rightarrow c=\frac{16}{3}$ <br/><br/>Now, $x=8$ <br/><br/>$8 f(8)=\frac{27}{3}-3+\frac{16}{3}=\frac{34}{3}$ <br/><br/>$12 f(8)=\frac{34}{3} \times \frac{12}{8}=17$