Let  $$\begin{aligned}3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\end{aligned}$$ Then what is $$8 \int_{1}^{2} f(x) \, dx$$ equal to?

Correct option is A

Solution:

​$$\textbf{Given:} \\3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1 \\\text{We need to find: } 8 \int_{1}^{2} f(x) \, dx \\\\\ \\\textbf{Step 1: Write two equations} \\\ \\3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1 \quad (1) \\\ \\3f\left(\frac{1}{x}\right) + f(x) = x + 1 \quad (2) \\\\\ \\\textbf{Step 2: Solve (1) and (2) by elimination} \\9f(x) + 3f\left(\frac{1}{x}\right) = 3 \left( \frac{1}{x} + 1 \right) \\-(f(x) + 3f\left(\frac{1}{x}\right) = x + 1) \\\\\ \\\text{Subtracting:} \\8f(x) = 3 \left( \frac{1}{x} + 1 \right) - (x + 1) \\\ \\8f(x) = \frac{3}{x} + 3 - x - 1 \\\ \\8f(x) = \frac{3}{x} + 2 - x \\\\\ \\\textbf{Step 3: Integrate both sides from 1 to 2} \\\ \\8 \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \left( \frac{3}{x} + 2 - x \right) \, dx \\\\\textbf{Step 4: Solve the integrals separately} \\\int_{1}^{2} \frac{3}{x} \, dx = 3 (\ln 2 - \ln 1) = 3 \ln 2 \\\ \\\int_{1}^{2} 2 \, dx = 2(2-1) = 2 \\\int_{1}^{2} x \, dx = \frac{x^2}{2} \Big|_{1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \\\\\ \\\textbf{Step 5: Combine all results} \\\ \\8 \int_{1}^{2} f(x) \, dx = 3 \ln 2 + 2 - \frac{3}{2} \\\ \\= 3 \ln 2 + \frac{1}{2} \\\\\ \\\textbf{Step 6: Express final result as log expression} \\\ \\8 \int_{1}^{2} f(x) \, dx = \ln 8 + \ln \sqrt{e} = \ln (8 \cdot \sqrt{e}) \\\\\ \\\textbf{Final Answer: } \boxed{\log 8 \sqrt{e}}$$​​