Indefinite Integration
Integration using ∫[f(x)+f'(x)]eˣ form
Grade 12
Question:
<p>We have \[I = \int\left\{\frac{(\log x - 1)}{1 + (\log x)^2}\right\}^2 dx\] Then \(I\) equals:</p>
<p>\(\dfrac{x}{1 + (\log x)^2} + C\)</p>
<p>\(\dfrac{\log x}{1 + (\log x)^2} + C\)</p>
<p>\(\dfrac{x^2}{1 + (\log x)^2} + C\)</p>
<p>\(\dfrac{x}{1 - (\log x)^2} + C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand can be decomposed using the derivative of arctan(log x), which is 1/(x(1+(log x)²)). Multiply and divide strategically to create a perfect differential form involving d(arctan(log x)).
<p><strong>Step 1:</strong> Recognize the structure. Note that d/dx[arctan(log x)] = 1/(x(1+(log x)²))</p><p><strong>Step 2:</strong> Let u = log x, so du = dx/x, thus dx = x·du = e^u·du</p><p><strong>Step 3:</strong> Rewrite the integrand:</p><p>I = ∫[(u-1)/(1+u²)]² · e^u du</p><p><strong>Step 4:</strong> Expand: [(u-1)/(1+u²)]² = (u-1)²/(1+u²)² = (u² - 2u + 1)/(1+u²)²</p><p><strong>Step 5:</strong> Split into parts and use the key observation that d/du[u/(1+u²)] = (1-u²)/(1+u²)² and d/du[arctan(u)] = 1/(1+u²)</p><p><strong>Step 6:</strong> After integration by parts and simplification:</p><p>I = e^u · [arctan(u) - u/(1+u²)] + C</p><p><strong>Step 7:</strong> Substitute back u = log x:</p><p>∴ I = x[arctan(log x) - (log x)/(1+(log x)²)] + C</p>
Correct Answer: A