Let $$5f(x) + 4f\left(\dfrac{1}{x}\right) = \dfrac{1}{x} + 3$$, $$x > 0$$. Then $$18\int_1^2 f(x)\,dx$$ is equal to

Given: $$5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3$$ ... (i)

Replace $$x$$ by $$\frac{1}{x}$$:

$$ 5f\left(\frac{1}{x}\right) + 4f(x) = x + 3 \quad \text{... (ii)} $$

Multiply (i) by 5 and (ii) by 4 and subtract:

$$ 25f(x) + 20f\left(\frac{1}{x}\right) - 20f\left(\frac{1}{x}\right) - 16f(x) = \frac{5}{x} + 15 - 4x - 12 $$

$$ 9f(x) = \frac{5}{x} - 4x + 3 $$

$$ f(x) = \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right) $$

Now compute:

$$ 18\int_1^2 f(x)\,dx = 2\int_1^2 \left(\frac{5}{x} - 4x + 3\right)dx $$

$$ = 2\left[5\ln x - 2x^2 + 3x\right]_1^2 $$

$$ = 2\left[(5\ln 2 - 8 + 6) - (0 - 2 + 3)\right] $$

$$ = 2\left[5\ln 2 - 2 - 1\right] = 2(5\ln 2 - 3) = 10\ln 2 - 6 $$

The correct answer is $$10\log_e 2 - 6$$.