Let a differentiable function $$f$$ satisfy $$f(x) + \int_3^x \dfrac{f(t)}{t} dt = \sqrt{x+1}$$, $$x \ge 3$$. Then $$12f(8)$$ is equal to:

Given: $$f(x) + \int_3^x \frac{f(t)}{t} dt = \sqrt{x+1}$$, $$x \geq 3$$

We start by noting that At $$x = 3$$: $$f(3) + 0 = \sqrt{4} = 2$$, so $$f(3) = 2$$.

Next, Differentiate both sides with respect to $$x$$:

$$ f'(x) + \frac{f(x)}{x} = \frac{1}{2\sqrt{x+1}} $$

This is a linear ODE. Integrating factor: $$\mu = e^{\int \frac{1}{x}dx} = x$$.

$$ \frac{d}{dx}[xf(x)] = \frac{x}{2\sqrt{x+1}} $$

From this, Integrate from 3 to $$x$$:

$$ xf(x) - 3f(3) = \int_3^x \frac{t}{2\sqrt{t+1}} dt $$

Substituting $$u = t + 1$$:

$$ \int \frac{t}{2\sqrt{t+1}} dt = \int \frac{u-1}{2\sqrt{u}} du = \frac{u^{3/2}}{3} - \sqrt{u} = \frac{(t+1)^{3/2}}{3} - \sqrt{t+1} $$

$$ xf(x) = 6 + \left[\frac{(x+1)^{3/2}}{3} - \sqrt{x+1}\right] - \left[\frac{8}{3} - 2\right] $$

$$ xf(x) = \frac{16}{3} + \frac{(x+1)^{3/2}}{3} - \sqrt{x+1} $$

Building on the above, At $$x = 8$$:

$$ 8f(8) = \frac{16}{3} + \frac{27}{3} - 3 = \frac{16}{3} + 9 - 3 = \frac{16}{3} + 6 = \frac{34}{3} $$

$$ 12f(8) = \frac{12 \times 34}{3 \times 8} = \frac{12 \times 34}{24} = 17 $$