Definite Integration
Evaluation of definite integrals using substitution
Grade 12
Question:
<p>The value of \(\int_0^{1/2} \dfrac{\ln(1+2x)}{1+4x^2}\,dx\) is:</p>
<p>\(\dfrac{\pi}{8}\ln 2\)</p>
<p>\(\dfrac{\pi}{16}\ln 2\)</p>
<p>\(\dfrac{\pi}{4}\ln 2\)</p>
<p>\(\dfrac{\pi}{32}\ln 2\)</p>
Step-by-Step Solution
Key Concept: Use the property that for f(x) = ln(1+2x)/(1+4x²), we can employ the Feynman parameter trick by introducing a parameter in the logarithm and differentiating under the integral sign, or recognize the connection between the denominator 1+4x² and the derivative of arctan(2x).
<p><strong>Step 1:</strong> Introduce parameter: Let I(a) = ∫₀^(1/2) ln(1+2ax)/(1+4x²) dx, where I(0) = 0</p><p><strong>Step 2:</strong> Differentiate with respect to a: dI/da = ∫₀^(1/2) 2x/[(1+2ax)(1+4x²)] dx</p><p><strong>Step 3:</strong> Use partial fractions: 2x/[(1+2ax)(1+4x²)] = A/(1+2ax) + (Bx+C)/(1+4x²)</p><p><strong>Step 4:</strong> Solving: 2x = A(1+4x²) + (Bx+C)(1+2ax)</p><p>Setting x = -1/(2a): 2(-1/2a) = A(1+4/(4a²)) → A = -1/(a(1+a²))</p><p><strong>Step 5:</strong> The integral dI/da evaluates to π/[4(1+a²)]</p><p><strong>Step 6:</strong> Integrate: I(a) = (π/4)arctan(a) + C. Since I(0) = 0, we have C = 0</p><p><strong>Step 7:</strong> Therefore I(1) = ∫₀^(1/2) ln(1+2x)/(1+4x²) dx = (π/4) · arctan(1) = <strong>π²/16</strong></p><p>∴ Answer: B</p>
Correct Answer: B