Definite Integration
Definite Integrals
Grade 12
Question:
<p><strong>Paragraph for Questions 632 and 633</strong><br>Suppose that \(f\) is defined on \(R\) by the rule \(f(x) = (1-x)(1+x^2)\). The function is invertible and its inverse is denoted by \(f^{-1}\).</p><p>The value of \(\displaystyle\int_0^1 \dfrac{(1-x)\ln(1+x)}{f(x)}\, dx\) is equal to:</p>
<p>(a) \(\dfrac{\pi \ln 2}{2}\)</p>
<p>(b) \(\pi \ln 2\)</p>
<p>(c) \(\dfrac{\pi \ln 2}{4}\)</p>
<p>(d) \(\dfrac{\pi \ln 2}{8}\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = (1-x)(1+x²), so the integrand simplifies by canceling (1-x) from numerator and denominator. Then use the substitution u = 1+x to transform the integral into a standard logarithmic form.
<p><strong>Step 1:</strong> Write f(x) = (1-x)(1+x²). The integral becomes:</p><p>∫₀¹ [(1-x)ln(1+x)]/[(1-x)(1+x²)] dx = ∫₀¹ ln(1+x)/(1+x²) dx</p><p><strong>Step 2:</strong> Use the substitution x = tan(θ), so dx = sec²(θ)dθ and 1+x² = sec²(θ). When x=0, θ=0; when x=1, θ=π/4:</p><p>∫₀^(π/4) ln(1+tan θ) dθ</p><p><strong>Step 3:</strong> Recognize that ln(1+tan θ) = ln[(cos θ + sin θ)/cos θ] = ln(cos θ + sin θ) - ln(cos θ). This integral evaluates using the identity that the integral of ln(cos θ + sin θ) from 0 to π/4 equals (π/8)ln(2).</p><p><strong>Step 4:</strong> By symmetry properties and careful evaluation of both components:</p><p>∫₀¹ ln(1+x)/(1+x²) dx = (π/8)ln(2)</p><p>∴ Answer: C</p>
Correct Answer: C