<p>Let \(f(x) = \int_0^x \ln(1+t^2)\, dt + x\ln(1+x^2)\). Then \(f''(0)\) equals:</p>
Step-by-Step Solution
Key Concept: Use Leibniz rule for differentiating integrals: d/dx[∫₀ˣ g(t)dt] = g(x). Then differentiate twice, applying the product rule to the second term.
<p><strong>Step 1:</strong> Find f'(x) using Leibniz rule and product rule.</p><p>f(x) = ∫₀ˣ ln(1+t²)dt + x·ln(1+x²)</p><p>f'(x) = ln(1+x²) + [ln(1+x²) + x·(2x)/(1+x²)]</p><p>f'(x) = 2ln(1+x²) + 2x²/(1+x²)</p><p><strong>Step 2:</strong> Find f''(x) by differentiating f'(x).</p><p>f''(x) = 2·(2x)/(1+x²) + d/dx[2x²/(1+x²)]</p><p>For the second term, use quotient rule:</p><p>d/dx[2x²/(1+x²)] = [4x(1+x²) - 2x²(2x)]/(1+x²)² = [4x + 4x³ - 4x³]/(1+x²)² = 4x/(1+x²)²</p><p>f''(x) = 4x/(1+x²) + 4x/(1+x²)²</p><p><strong>Step 3:</strong> Evaluate at x=0.</p><p>f''(0) = 4(0)/(1+0) + 4(0)/(1+0)² = 0 + 0 = <strong>0</strong></p><p>∴ Answer: A</p>
Correct Answer: A