Indefinite Integration
Integration by Partial Fractions
Grade 12
Question:
<p>Evaluate \(\int \frac{1 - x^2}{x(1 - 2x)} dx\)</p>
<p>(a) \(\frac{x}{3} + \log x - \frac{3}{4} \log|x - 1| + C\)</p>
<p>(b) \(\frac{x}{2} + \log x + \frac{1}{4} \log|x - 1| + C\)</p>
<p>(c) \(\frac{x}{2} + \log x - \frac{3}{4} \log|2x - 1| + C\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Decompose the rational function using partial fractions and integrate each term separately.
<p><strong>Step 1:</strong> Let $I = \int \frac{1 - x^2}{x(1 - 2x)} dx$</p><p><strong>Step 2:</strong> Use partial fractions: $\frac{1 - x^2}{x(1 - 2x)} = \frac{A}{x} + \frac{B}{2x - 1}$</p><p><strong>Step 3:</strong> Multiplying both sides by $x(1 - 2x)$: $1 - x^2 = A(1 - 2x) + Bx$</p><p><strong>Step 4:</strong> Substituting $x = 0$: $1 = A$</p><p><strong>Step 5:</strong> Substituting $x = \frac{1}{2}$: $\frac{3}{4} = \frac{B}{2}$ gives $B = -\frac{3}{2}$</p><p><strong>Step 6:</strong> $I = \int \frac{1}{x} dx - \frac{3}{2} \int \frac{1}{2x - 1} dx = \log x - \frac{3}{4} \log|2x - 1| + C$</p><p><strong>Step 7:</strong> Including the polynomial part: $I = \frac{x}{2} + \log x - \frac{3}{4} \log|2x - 1| + C$</p><p>∴ Answer is (c).</p>
Correct Answer: C