Indefinite Integration
Integration using sum of squares
Grade 12
Question:
<p>Evaluate <span class="math">\(\int \frac{dx}{9x^2 + 6x + 5}\)</span></p>
<p>(A) <span class="math">\(\frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right) + C\)</span></p>
<p>(B) <span class="math">\(\frac{1}{2}\tan^{-1}\left(\frac{3x+1}{2}\right) + C\)</span></p>
<p>(C) <span class="math">\(\frac{1}{6}\cot^{-1}\left(\frac{3x+1}{2}\right) + C\)</span></p>
<p>(D) <span class="math">\(\frac{1}{2}\cot^{-1}\left(\frac{3x+1}{2}\right) + C\)</span></p>
Step-by-Step Solution
Key Concept: Convert the quadratic into sum of squares (perfect square + constant), then apply the standard inverse tangent integral formula.
<p><strong>Step 1:</strong> Express the denominator as a sum of squares.</p><p><span class="math">$9x^2 + 6x + 5 = 9x^2 + 6x + 1 + 4 = (3x+1)^2 + 4$</span></p><p><strong>Step 2:</strong> Apply the standard formula <span class="math">$\int \frac{dt}{t^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right) + C$</span></p><p><strong>Step 3:</strong> Let <span class="math">$t = 3x + 1$</span>, then <span class="math">$dt = 3dx$</span></p><p><span class="math">$\int \frac{dx}{(3x+1)^2 + 4} = \frac{1}{3}\int \frac{dt}{t^2 + 4} = \frac{1}{3} \cdot \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) + C = \frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right) + C$</span></p><p>∴ Answer is A.</p>
Correct Answer: A