Indefinite Integration
Special Functions Integration
Grade 12
Question:
<p>If <span class="math">\int \frac{\log(x + \sqrt{1 + x^2})}{1 + x}dx = g \circ f(x) + \text{constant}</span>, then (JEE Main 2016)</p>
<p>(a) <span class="math">f(x) = \log(x + \sqrt{x^2 + 1})</span></p>
<p>(b) <span class="math">f(x) = \log(x + \sqrt{x + 1})</span> and <span class="math">g(x) = x</span></p>
<p>(c) <span class="math">f(x) = \log(x + \sqrt{x^2 + 1})</span> and <span class="math">g(x) = \frac{x^2}{2}</span></p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Recognize function composition in the antiderivative; use substitution to simplify logarithmic integrals.
<p><strong>Step 1:</strong> Let <span class="math">u = \log(x + \sqrt{1 + x^2})</span>, then <span class="math">du = \frac{1}{\sqrt{1+x^2}}dx</span></p><p><strong>Step 2:</strong> Use substitution and recognize the composition of functions in the result.</p><p><strong>Step 3:</strong> The integral evaluates to <span class="math">\frac{[\log(x + \sqrt{1 + x^2})]^2}{2} + C</span>, which is <span class="math">g \circ f(x)</span> where <span class="math">f(x) = \log(x + \sqrt{x^2+1})</span> and <span class="math">g(x) = \frac{x^2}{2}</span></p><p>∴ Answer is (c).</p>
Correct Answer: c