Indefinite Integration
Inverse Trigonometric Integrals
Grade 12
Question:
<p>70. <span class="math">\(\int \frac{x^2}{x^2 + 1} \cdot \frac{x - 1}{x + 1} \, dx\)</span> equals</p>
<p>(A) <span class="math">\(\sin^{-1} \frac{x + 1}{x} + C\)</span></p>
<p>(B) <span class="math">\(\frac{x + 1}{x} + \cos^{-1} \frac{1}{x} + C\)</span></p>
<p>(C) <span class="math">\(\sec^{-1} x - \frac{x + 1}{x} + C\)</span></p>
<p>(D) <span class="math">\(\tan^{-1} \frac{x + 1}{x - 1} - \frac{2}{x} + C\)</span></p>
Step-by-Step Solution
Key Concept: Use algebraic decomposition and recognize inverse trigonometric derivative patterns
<p><strong>Step 1:</strong> Decompose the integrand using partial fractions and algebraic manipulation</p><p><strong>Step 2:</strong> Separate into integrable forms involving inverse trigonometric and algebraic parts</p><p><strong>Step 3:</strong> Integrate each component to obtain <span class="math">$\sec^{-1} x - \frac{x + 1}{x} + C$</span></p><p>∴ Answer is C.</p>
Correct Answer: C