Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>If <span>\(\int \frac{x+1}{\sqrt{2x-1}}dx = f(x)\sqrt{2x-1} + C\)</span>, where <em>C</em> is a constant of integration, then <em>f(x)</em> is equal to:</p>
<p>\(\frac{1}{3}(x+1)\)</p>
<p>\(\frac{2}{3}(x+2)\)</p>
<p>\(\frac{2}{3}(x-4)\)</p>
<p>\(\frac{1}{3}(x+4)\)</p>
Step-by-Step Solution
Key Concept: Use substitution u = 2x - 1 to convert the integrand, then express the result in the required form f(x)√(2x-1) to identify f(x) by comparing coefficients.
<p><strong>Step 1:</strong> Let u = 2x - 1, so du = 2dx, and x = (u+1)/2</p><p><strong>Step 2:</strong> Rewrite the integral:</p><p>∫ (x+1)/√(2x-1) dx = ∫ [(u+1)/2 + 1]/√u · (du/2)</p><p>= ∫ [(u+3)/2]/√u · (du/2) = (1/4)∫ (u+3)/√u du</p><p><strong>Step 3:</strong> Expand: (1/4)∫ (u^(1/2) + 3u^(-1/2)) du</p><p>= (1/4)[2u^(3/2)/3 + 6u^(1/2)] + C</p><p>= (1/4) · (2/3)u^(3/2) + (1/4) · 6u^(1/2) + C</p><p>= (u^(3/2)/6) + (3u^(1/2)/2) + C</p><p><strong>Step 4:</strong> Factor out √u = √(2x-1):</p><p>= √(2x-1)[u/6 + 3/2] + C</p><p>= √(2x-1)[(2x-1)/6 + 3/2] + C</p><p>= √(2x-1)[(2x-1)/6 + 9/6] + C</p><p>= √(2x-1)[(2x+8)/6] + C</p><p><strong>Step 5:</strong> Therefore f(x) = (2x+8)/6 = <strong>(x+4)/3</strong></p><p>∴ Answer: D</p>
Correct Answer: D