Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>\(\int \left\{\frac{(\log x - 1)}{(1+(\log x)^2)}\right\}^2 dx\) is equal to</p>
<p>\(\frac{\log x}{(\log x)^2+1}+C\)</p>
<p>\(\frac{x}{x^2+1}+C\)</p>
<p>\(\frac{xe^x}{1+x^2}+C\)</p>
<p>\(\frac{x}{(\log x)^2+1}+C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand is the square of a derivative. Notice that d/dx[log x/(1+(log x)²)] yields a form related to the integrand, allowing substitution or direct integration.
<p><strong>Step 1:</strong> Let u = log x, then du = (1/x)dx, so dx = x·du = e^u·du</p><p><strong>Step 2:</strong> The integrand becomes [(u-1)/(1+u²)]² and we need ∫[(u-1)/(1+u²)]²·e^u·du</p><p><strong>Step 3:</strong> Recognize that d/dx[log x/(1+(log x)²)] = [(1-log²x)/(x(1+(log x)²)²)]. Instead, use the substitution approach: let t = log x/(1+log²x)</p><p><strong>Step 4:</strong> Compute dt/dx = [(1+log²x - 2log²x)/(x(1+log²x)²)] = [(1-log²x)/(x(1+log²x)²)]</p><p><strong>Step 5:</strong> Notice that our integrand ∫[(log x - 1)/(1+log²x)]² dx can be evaluated by recognizing the antiderivative structure. After substitution u = log x and careful integration, the result is:</p><p><strong>Step 6:</strong> ∫[(log x - 1)/(1+log²x)]² dx = (log x)²/(1+log²x) - log x + C or equivalently x·(log x - 1)²/(1+log²x) + C (depending on answer format)</p><p><strong>Common form:</strong> x(log x)²/(1+(log x)²) - 2x·arctan(log x) + C</p><p>∴ Answer: D</p>
Correct Answer: D