<p>Let <br/>
\[ I = \int_{0}^{7} \frac{x\log x}{(1+x^2)^2}\, dx \]
Find the value of \(I\).</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be split using substitution properties: rewrite the numerator strategically and use the fact that f(x) + f(-x) relationships often yield zero for symmetric intervals when properly transformed.
<p><strong>Step 1:</strong> Apply substitution u = 1/x, so x = 1/u, dx = -du/u²</p><p>When x = 0⁺, u → ∞; when x = 7, u = 1/7</p><p>I = ∫₀⁷ (x log x)/(1+x²)² dx = ∫∞^(1/7) [(1/u)log(1/u)]/(1+1/u²)² · (-du/u²)</p><p><strong>Step 2:</strong> Simplify the integrand:</p><p>= ∫_(1/7)^∞ [(1/u)(-log u)]/[(u²+1)²/u⁴] · (du/u²)</p><p>= ∫_(1/7)^∞ [(-log u) · u⁴]/[u · (u²+1)² · u²] du</p><p>= ∫_(1/7)^∞ [(-log u) · u]/[(u²+1)²] du</p><p><strong>Step 3:</strong> Change limits back: letting v = 1/u gives</p><p>∫₀⁷ [(-log(1/v)) · (1/v)]/[(1/v²+1)²] · (-dv/v²) = -∫₀⁷ (v log v)/(1+v²)² dv = -I</p><p><strong>Step 4:</strong> From Step 3: I = -I, therefore 2I = 0</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0