Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>If \(f\left(\dfrac{x-4}{x+2}\right) = 2x + 1\), \((x \in \mathbb{R} - \{1, -2\})\), then \(\int f(x)\, dx\) is equal to (where \(C\) is a constant of integration):</p>
<p>\(12\ln|1-x| + 3x + C\)</p>
<p>\(-12\ln|1-x| - 3x + C\)</p>
<p>\(12\ln|1-x| - 3x + C\)</p>
<p>\(-12\ln|1-x| + 3x + C\)</p>
Step-by-Step Solution
Key Concept: Find f(x) by substituting u = (x-4)/(x+2), then solve for x in terms of u to express the given function in standard form before integrating.
<p><strong>Step 1:</strong> Let u = (x-4)/(x+2). We need to express x in terms of u.</p><p>u(x+2) = x-4</p><p>ux + 2u = x - 4</p><p>ux - x = -4 - 2u</p><p>x(u-1) = -4 - 2u</p><p>x = (-4-2u)/(u-1) = (2u+4)/(1-u)</p><p><strong>Step 2:</strong> Since f(u) = 2x + 1, substitute the expression for x:</p><p>f(u) = 2·(2u+4)/(1-u) + 1</p><p>f(u) = (4u+8)/(1-u) + 1</p><p>f(u) = (4u+8+1-u)/(1-u)</p><p>f(u) = (3u+9)/(1-u)</p><p><strong>Step 3:</strong> Therefore f(x) = (3x+9)/(1-x). Rewrite using polynomial division:</p><p>f(x) = (3x+9)/(1-x) = -3 - 6/(1-x)</p><p><strong>Step 4:</strong> Integrate:</p><p>∫f(x)dx = ∫[-3 - 6/(1-x)]dx</p><p>= -3x - 6·ln|1-x|·(-1) + C</p><p>= -3x + 6ln|1-x| + C</p><p>∴ Answer: B</p>
Correct Answer: B