Indefinite Integration
Reduction formula / Integration by substitution
Grade 12
Question:
<p>Let <br>\[I = \int \frac{dx}{(x^2 - 2x + 10)^2}\]<br>If \(I = A\left[\tan^{-1}\left(\frac{x-1}{3}\right) + \frac{f(x)}{x^2 - 2x + 10}\right] + C\), then \(A\) and \(f(x)\) are:</p>
<p>\(A = \frac{1}{54}\) and \(f(x) = 3(x-1)\)</p>
<p>\(A = \frac{1}{54}\) and \(f(x) = -3(x-1)\)</p>
<p>\(A = \frac{1}{27}\) and \(f(x) = 3(x-1)\)</p>
<p>\(A = \frac{1}{27}\) and \(f(x) = -3(x-1)\)</p>
Step-by-Step Solution
Key Concept: Complete the square to convert the denominator to (u² + a²)² form, then use the standard reduction formula: ∫du/(u² + a²)ⁿ = (u/(2(n-1)a²(u² + a²)ⁿ⁻¹)) + ((2n-3)/(2(n-1)a²))∫du/(u² + a²)ⁿ⁻¹. This produces both an inverse tangent term and a rational function term.
<p><strong>Step 1:</strong> Complete the square in the denominator.</p><p>x² - 2x + 10 = (x-1)² + 9</p><p>Let u = x - 1, so du = dx</p><p>I = ∫du/((u² + 9)²)</p><p><strong>Step 2:</strong> Apply the reduction formula for ∫du/(u² + a²)ⁿ with n = 2 and a² = 9:</p><p>∫du/(u² + 9)² = [u/(2·1·9(u² + 9))] + (1/(2·1·9))∫du/(u² + 9)</p><p>= [u/(18(u² + 9))] + (1/18)·(1/3)tan⁻¹(u/3)</p><p>= [u/(18(u² + 9))] + (1/54)tan⁻¹(u/3)</p><p><strong>Step 3:</strong> Substitute back u = x - 1:</p><p>I = [(x-1)/(18(x² - 2x + 10))] + (1/54)tan⁻¹((x-1)/3)</p><p><strong>Step 4:</strong> Rewrite in the given form:</p><p>I = (1/54)[tan⁻¹((x-1)/3) + (3(x-1))/(x² - 2x + 10)] + C</p><p>I = (1/54)[tan⁻¹((x-1)/3) + (3x-3)/(x² - 2x + 10)] + C</p><p><strong>Therefore:</strong> A = 1/54 and f(x) = 3x - 3 or f(x) = 3(x-1)</p>
Correct Answer: A