Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Given \(I = \displaystyle\int \frac{2x^3 - 1}{x^4 + x}\,dx\). Then \(I\) equals:</p>
<p>\(\log_e\left|\dfrac{x^3+1}{x}\right| + C\)</p>
<p>\(\log_e\left|x^2 + \dfrac{1}{x}\right| + C\)</p>
<p>\(\log_e\left|x^3 - 1\right| + C\)</p>
<p>\(\log_e\left|x^4 + x\right| + C\)</p>
Step-by-Step Solution
Key Concept: Decompose the numerator strategically: write 2x³ - 1 = (1/2)·d/dx(x⁴ + x) + remainder, which splits the integral into a logarithmic part and a rational function.
<p><strong>Step 1:</strong> Recognize that d/dx(x⁴ + x) = 4x³ + 1</p><p><strong>Step 2:</strong> Rewrite the numerator as: 2x³ - 1 = (1/2)(4x³ + 1) - (3/2)</p><p>So: I = ∫[(1/2)(4x³ + 1) - (3/2)]/(x⁴ + x) dx</p><p><strong>Step 3:</strong> Split into two integrals:</p><p>I = (1/2)∫(4x³ + 1)/(x⁴ + x) dx - (3/2)∫1/(x⁴ + x) dx</p><p><strong>Step 4:</strong> First integral: (1/2)ln|x⁴ + x| + C₁</p><p><strong>Step 5:</strong> For second integral, factor: x⁴ + x = x(x³ + 1) = x(x + 1)(x² - x + 1)</p><p>Use partial fractions: 1/[x(x + 1)(x² - x + 1)] and integrate</p><p><strong>Step 6:</strong> After partial fraction decomposition and integration:</p><p>-(3/2)[ln|x| - ln|x + 1|/2 + ...] = -(3/2)ln|x| + (3/4)ln|x + 1| + constant terms</p><p><strong>Final Answer:</strong> I = (1/2)ln|x⁴ + x| - (3/2)ln|x| + (3/4)ln|x + 1| + C = (1/2)ln|x(x³ + 1)| - (3/2)ln|x| + (3/4)ln|x + 1| + C</p><p>Which simplifies to: <strong>(1/2)ln|x³ + 1| + (1/4)ln|x + 1| + C</strong> or equivalent form</p>
Correct Answer: A