<p>If \(\displaystyle\int_1^2 \dfrac{dx}{(x^2-2x+4)^{3/2}} = \dfrac{k}{k+5}\), then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Complete the square in the denominator to get (x-1)² + 3, then use the substitution x - 1 = √3 tan(θ) to convert to a trigonometric integral that evaluates to a standard form.
<p><strong>Step 1:</strong> Complete the square: x² - 2x + 4 = (x-1)² + 3</p><p><strong>Step 2:</strong> Rewrite the integral: ∫₁² dx/[(x-1)² + 3]^(3/2)</p><p><strong>Step 3:</strong> Substitute x - 1 = √3 tan(θ), so dx = √3 sec²(θ) dθ</p><p>When x = 1: θ = 0; When x = 2: tan(θ) = 1/√3, so θ = π/6</p><p><strong>Step 4:</strong> The denominator becomes [(x-1)² + 3]^(3/2) = [3 tan²(θ) + 3]^(3/2) = 3√3 sec³(θ)</p><p><strong>Step 5:</strong> The integral becomes: ∫₀^(π/6) [√3 sec²(θ)]/[3√3 sec³(θ)] dθ = (1/3)∫₀^(π/6) cos(θ) dθ</p><p><strong>Step 6:</strong> Evaluate: (1/3)[sin(θ)]₀^(π/6) = (1/3) · (1/2) = 1/6</p><p><strong>Step 7:</strong> Given that 1/6 = k/(k+5), solve: k + 5 = 6k, so 5 = 5k, therefore k = 1</p><p>∴ Answer: D (k = 1)</p>
Correct Answer: D