Indefinite Integration
Integration by parts / substitution
Grade 12
Question:
<p>If \(\displaystyle\int x^{26}(x-1)^{17}(5x-3)\, dx = \dfrac{x^{27}(x-1)^{18}}{k} + C\), where \(C\) is constant of integration, then the value of \(k\) is:</p>
<p>(a) 3</p>
<p>(b) 6</p>
<p>(c) 9</p>
<p>(d) 12</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand x^26(x-1)^17(5x-3) is the derivative of x^27(x-1)^18 up to a constant factor. Use the product rule in reverse: if d/dx[x^27(x-1)^18] = x^27·18(x-1)^17 + (x-1)^18·27x^26, simplify to find the constant k.
<p><strong>Step 1:</strong> Assume ∫ x^26(x-1)^17(5x-3) dx = x^27(x-1)^18/k + C. Differentiate both sides with respect to x.</p><p><strong>Step 2:</strong> Using the product rule on the right side: d/dx[x^27(x-1)^18] = x^27·18(x-1)^17 + (x-1)^18·27x^26</p><p><strong>Step 3:</strong> Factor out x^26(x-1)^17: = x^26(x-1)^17[18x + 27(x-1)] = x^26(x-1)^17[18x + 27x - 27] = x^26(x-1)^17[45x - 27]</p><p><strong>Step 4:</strong> Factor out 9: = 9·x^26(x-1)^17[5x - 3]</p><p><strong>Step 5:</strong> Therefore: d/dx[x^27(x-1)^18/k] = x^26(x-1)^17(5x-3) gives us 9/k = 1</p><p>∴ <strong>k = 9</strong></p>
Correct Answer: C