<p>Let \(F(x) = f(x) + f\!\left(\dfrac{1}{x}\right)\), where \(f(x) = \int_{1}^{x} \dfrac{\log t}{1+t} \, dt\). Then \(F(e)\) equals</p>
Step-by-Step Solution
Key Concept: Use the substitution property of definite integrals: when computing f(1/x), substitute t = 1/u to relate it back to f(x), revealing that F(x) = f(x) + f(1/x) becomes a constant that can be evaluated at any convenient point.
<p><strong>Step 1:</strong> Find f(1/x) by substituting t = 1/u in the integral:</p><p>f(1/x) = ∫₁^(1/x) [log t/(1+t)] dt</p><p>Let t = 1/u, then dt = -du/u². When t = 1, u = 1; when t = 1/x, u = x.</p><p><strong>Step 2:</strong> Substitute:</p><p>f(1/x) = ∫₁ˣ [log(1/u)/(1+1/u)] · (-du/u²) = ∫₁ˣ [-log u/(1+u)] · (du/u²) · u²/(1+u)</p><p>= ∫₁ˣ [-log u/(1+u)] du = -f(x)</p><p><strong>Step 3:</strong> Therefore:</p><p>F(x) = f(x) + f(1/x) = f(x) - f(x) = 0</p><p><strong>Step 4:</strong> Thus:</p><p>F(e) = 0</p><p>∴ Answer: A</p>
Correct Answer: A