<p>The integral \(\displaystyle\int_2^4 \dfrac{\log x^2}{\log x^2 + \log(36-12x+x^2)}\,dx\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the property that ∫[a to b] f(x)dx = ∫[a to b] f(a+b-x)dx, then add the original and transformed integrals to simplify the denominator. Notice that 36-12x+x² = (x-6)², so log(36-12x+x²) = 2log|x-6|.
<p><strong>Step 1:</strong> Recognize the form and apply King's property. Let I = ∫₂⁴ [log(x²)]/[log(x²) + log(36-12x+x²)] dx</p><p><strong>Step 2:</strong> Substitute x → 6-x (since a+b=2+4=6). When x=2, new variable=4; when x=4, new variable=2.</p><p>I = ∫₂⁴ [log((6-x)²)]/[log((6-x)²) + log(36-12(6-x)+(6-x)²)] dx</p><p><strong>Step 3:</strong> Simplify the denominator. Note that 36-12(6-x)+(6-x)² = 36-72+12x+36-12x+x² = x². So:</p><p>I = ∫₂⁴ [2log(6-x)]/[2log(6-x) + 2log(x)] dx = ∫₂⁴ [log(6-x)]/[log(6-x) + log(x)] dx</p><p><strong>Step 4:</strong> Add the original integral and this result:</p><p>2I = ∫₂⁴ [log(x²) + log(6-x)]/[log(x²) + log(36-12x+x²)] dx = ∫₂⁴ [2log(x) + log(6-x)]/[2log(x) + 2log(6-x)] dx</p><p><strong>Step 5:</strong> This simplifies to 2I = ∫₂⁴ 1 dx = 2, therefore I = 1</p><p>∴ Answer: B</p>
Correct Answer: B