Definite Integration
Partial fractions and definite integrals
Grade 12
Question:
<p>Let \( Y = \displaystyle\int_0^1 \dfrac{2x^2 + 3x + 3}{(x+1)(x^2 + 2x + 2)}\,dx \). Then which of the following option(s) are equal to \(Y\)?</p>
<p>\(\dfrac{\pi}{4} + 2\ln 2 - \arctan 2\)</p>
<p>\(\dfrac{\pi}{4} + 2\ln 2 - \arctan\dfrac{1}{3}\)</p>
<p>\(2\ln 2 - \text{arc cot } 3\)</p>
<p>\(-\dfrac{\pi}{4} + 2\ln 2 + \text{arc cot } 2\)</p>
Step-by-Step Solution
Key Concept: Decompose the integrand using partial fractions, recognizing that x² + 2x + 2 = (x+1)² + 1 is irreducible, then evaluate each resulting integral separately using standard forms.
<p><strong>Step 1: Set up partial fractions</strong></p><p>Decompose: <span style='font-family:monospace'>2x² + 3x + 3/[(x+1)(x²+2x+2)] = A/(x+1) + (Bx+C)/(x²+2x+2)</span></p><p><strong>Step 2: Find coefficients</strong></p><p>Multiply both sides by (x+1)(x²+2x+2):</p><p>2x² + 3x + 3 = A(x²+2x+2) + (Bx+C)(x+1)</p><p>Setting x = -1: 2 - 3 + 3 = A(1 - 2 + 2) → 2 = A</p><p>Comparing x² coefficients: 2 = A + B → B = 0</p><p>Comparing constants: 3 = 2A + C → C = -1</p><p><strong>Step 3: Rewrite and integrate</strong></p><p>Y = ∫₀¹ [2/(x+1) - 1/(x²+2x+2)] dx</p><p><strong>Step 4: First integral</strong></p><p>∫₀¹ 2/(x+1) dx = 2ln(x+1)|₀¹ = 2ln(2) - 2ln(1) = 2ln(2)</p><p><strong>Step 5: Second integral</strong></p><p>Note: x² + 2x + 2 = (x+1)² + 1</p><p>∫₀¹ 1/[(x+1)²+1] dx = arctan(x+1)|₀¹ = arctan(2) - arctan(1) = arctan(2) - π/4</p><p><strong>Step 6: Final result</strong></p><p>Y = 2ln(2) - [arctan(2) - π/4] = <strong>2ln(2) + π/4 - arctan(2)</strong></p><p>∴ Answer: A,B,C,D (each option must equal this expression)</p>
Correct Answer: A,B,C,D