<p>Let \(f(x) = \int_1^x \dfrac{\log t}{1+t}\,dt\). Then \(F(e) = f(e) + f\!\left(\dfrac{1}{e}\right)\) equals:</p>
Step-by-Step Solution
Key Concept: Use the property that f(1/x) can be related to f(x) through substitution u = 1/t, which converts the integral limits and reveals a symmetry relationship between f(e) and f(1/e).
<p><strong>Step 1:</strong> Write out the definitions:</p><p>f(e) = ∫₁ᵉ (log t)/(1+t) dt</p><p>f(1/e) = ∫₁^(1/e) (log t)/(1+t) dt</p><p><strong>Step 2:</strong> For f(1/e), reverse limits and use substitution u = 1/t, so t = 1/u, dt = -du/u²:</p><p>f(1/e) = -∫ₑ¹ (log(1/u))/(1+1/u) · (du/u²) = -∫ₑ¹ (-log u)/((u+1)/u) · (du/u²)</p><p>= ∫ₑ¹ (log u · u)/(u+1) · (du/u²) = ∫ₑ¹ (log u)/(u+1) du = -∫₁ᵉ (log u)/(u+1) du</p><p><strong>Step 3:</strong> Therefore:</p><p>f(e) + f(1/e) = ∫₁ᵉ (log t)/(1+t) dt - ∫₁ᵉ (log t)/(1+t) dt = 0</p><p>∴ F(e) = <strong>0</strong></p>
Correct Answer: A