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Step 1: Understand the given equation and find an initial condition. The given equation is an integral equation: $$f(x)+\int_3^x\dfrac{f(t)}{t}\,dt=\sqrt{x+1}$$ This equation holds for $x \geq 3$. We can find the value of $f(3)$ by substituting $x=3$ into the equation. The integral term will vanish as the upper and lower limits will be the same. $$f(3)+\int_3^3\dfrac{f(t)}{t}\,dt=\sqrt{3+1}$$ $$f(3)+0=\sqrt{4}$$ $$f(3)=2$$ Step 2: Differentiate the integral equation with respect to $x$. To transform the integral equation into a differential equation, we differentiate both sides with respect to $x$. We use the Leibniz integral rule for differentiating the integral term: $\frac{d}{dx} \int_{a}^{x} g(t) dt = g(x)$. Differentiating the left side: $$\frac{d}{dx}\left(f(x)+\int_3^x\dfrac{f(t)}{t}\,dt\right) = f'(x) + \dfrac{f(x)}{x}$$ Differentiating the right side: $$\frac{d}{dx}\left(\sqrt{x+1}\right) = \frac{d}{dx}\left((x+1)^{1/2}\right) = \frac{1}{2}(x+1)^{-1/2} = \frac{1}{2\sqrt{x+1}}$$ Equating the derivatives, we get a first-order linear differential equation: $$f'(x) + \dfrac{f(x)}{x} = \frac{1}{2\sqrt{x+1}}$$ Step 3: Solve the first-order linear differential equation. The differential equation is of the form $y' + P(x)y = Q(x)$, where $y=f(x)$, $P(x)=\frac{1}{x}$, and $Q(x)=\frac{1}{2\sqrt{x+1}}$. First, calculate the integrating factor (IF): $$\text{IF} = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|}$$ Since $x \geq 3$, we have $|x|=x$. So, $$\text{IF} = x$$ Multiply the differential equation by the integrating factor: $$x\left(f'(x) + \dfrac{f(x)}{x}\right) = x\left(\frac{1}{2\sqrt{x+1}}\right)$$ $$x f'(x) + f(x) = \frac{x}{2\sqrt{x+1}}$$ The left side is the derivative of the product $(x f(x))$: $$\frac{d}{dx}(x f(x)) = \frac{x}{2\sqrt{x+1}}$$ Step 4: Integrate both sides and find the constant of integration. Integrate both sides with respect to $x$: $$x f(x) = \int \frac{x}{2\sqrt{x+1}} dx$$ To evaluate the integral, let $u = x+1$. Then $x = u-1$ and $du = dx$. $$\int \frac{u-1}{2\sqrt{u}} du = \frac{1}{2} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) du$$ $$= \frac{1}{2} \int (u^{1/2} - u^{-1/2}) du$$ $$= \frac{1}{2} \left(\frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2}\right) + C$$ $$= \frac{1}{3}u^{3/2} - u^{1/2} + C$$ Substitute $u=x+1$ back: $$x f(x) = \frac{1}{3}(x+1)^{3/2} - (x+1)^{1/2} + C$$ $$x f(x) = (x+1)^{1/2} \left(\frac{1}{3}(x+1) - 1\right) + C$$ $$x f(x) = \sqrt{x+1} \left(\frac{x+1-3}{3}\right) + C$$ $$x f(x) = \frac{(x-2)\sqrt{x+1}}{3} + C$$ Now, use the initial condition $f(3)=2$ (from Step 1) to find $C$: $$3 f(3) = \frac{(3-2)\sqrt{3+1}}{3} + C$$ $$3(2) = \frac{(1)\sqrt{4}}{3} + C$$ $$6 = \frac{2}{3} + C$$ $$C = 6 - \frac{2}{3} = \frac{18-2}{3} = \frac{16}{3}$$ So, the expression for $x f(x)$ is: $$x f(x) = \frac{(x-2)\sqrt{x+1}}{3} + \frac{16}{3}$$ Dividing by $x$, we get $f(x)$: $$f(x) = \frac{(x-2)\sqrt{x+1}}{3x} + \frac{16}{3x}$$ Step 5: Calculate $12f(8)$. Substitute $x=8$ into the expression for $f(x)$: $$f(8) = \frac{(8-2)\sqrt{8+1}}{3(8)} + \frac{16}{3(8)}$$ $$f(8) = \frac{6\sqrt{9}}{24} + \frac{16}{24}$$ $$f(8) = \frac{6 \cdot 3}{24} + \frac{16}{24}$$ $$f(8) = \frac{18}{24} + \frac{16}{24}$$ $$f(8) = \frac{34}{24} = \frac{17}{12}$$ Finally, multiply by 12 to find $12f(8)$: $$12f(8) = 12 \times \frac{17}{12} = 17$$ The final answer is $\boxed{17}$.