Indefinite Integration
Integration of Irrational Functions
Grade 12
Question:
<p>If \(\int \dfrac{2x+5}{\sqrt{7-6x-x^2}}\, dx = A\sqrt{7-6x-x^2} + B\sin^{-1}\left(\dfrac{x+3}{4}\right) + C\), where \(C\) is a constant of integration, then the values of \(A\) and \(B\) are:</p>
<p>\(A = -2,\ B = 1\)</p>
<p>\(A = -2,\ B = -1\)</p>
<p>\(A = 2,\ B = -1\)</p>
<p>\(A = 2,\ B = 1\)</p>
Step-by-Step Solution
Key Concept: Split the numerator into two parts: one proportional to the derivative of the expression under the square root (giving a direct antiderivative), and one constant (giving an inverse sine form). Complete the square in the denominator to match standard integral forms.
<p><strong>Step 1: Decompose the numerator</strong></p><p>The derivative of (7-6x-x²) is d/dx(7-6x-x²) = -6-2x. We need to express (2x+5) in terms of this derivative:</p><p>2x+5 = -1·(-6-2x) + (-1) = -1·d/dx(7-6x-x²) - 1</p><p>So: ∫(2x+5)/√(7-6x-x²) dx = -∫(-6-2x)/√(7-6x-x²) dx - ∫1/√(7-6x-x²) dx</p><p><strong>Step 2: Evaluate first integral (substitution type)</strong></p><p>Let u = 7-6x-x², then du = (-6-2x)dx</p><p>∫(-6-2x)/√(7-6x-x²) dx = ∫du/√u = 2√u = 2√(7-6x-x²)</p><p>So the first part contributes: -2√(7-6x-x²)</p><p><strong>Step 3: Evaluate second integral (inverse sine form)</strong></p><p>Complete the square: 7-6x-x² = -(x²+6x-7) = -(x²+6x+9-16) = -(x+3)² + 16 = 16-(x+3)²</p><p>∫1/√(16-(x+3)²) dx = sin⁻¹((x+3)/4)</p><p><strong>Step 4: Combine results</strong></p><p>∫(2x+5)/√(7-6x-x²) dx = -2√(7-6x-x²) - sin⁻¹((x+3)/4) + C</p><p>Comparing with A√(7-6x-x²) + B·sin⁻¹((x+3)/4) + C:</p><p><strong>A = -2 and B = -1</strong></p>
Correct Answer: B