Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Given<br>\[ \int \frac{dx}{x^3(1+x^6)^{2/3}} = x\, f(x)\cdot(1+x^6)^{1/3} + C \]<br>find \(f(x)\).</p>
<p>\(f(x) = \dfrac{1}{2x^3}\)</p>
<p>\(f(x) = \dfrac{-1}{2x^3}\)</p>
<p>\(f(x) = \dfrac{1}{3x^3}\)</p>
<p>\(f(x) = \dfrac{-1}{3x^3}\)</p>
Step-by-Step Solution
Key Concept: Use substitution u = 1 + x⁶ to transform the integral, then recognize that the result must match the form xf(x)(1+x⁶)^(1/3) by comparing coefficients after differentiation or algebraic manipulation.
<p><strong>Step 1:</strong> Let u = 1 + x⁶, then du = 6x⁵ dx, so x⁵ dx = du/6</p><p><strong>Step 2:</strong> Rewrite the integrand: ∫ dx/[x³(1+x⁶)^(2/3)] = ∫ 1/[x³u^(2/3)] dx</p><p>Since u = 1 + x⁶, we have x⁶ = u - 1, so x³ = (u-1)^(1/2) [considering positive real part]</p><p><strong>Step 3:</strong> Express: ∫ dx/[x³(1+x⁶)^(2/3)] = ∫ (1/x⁵) · (x²/u^(2/3)) dx = (1/6) ∫ (x²/u^(2/3)) · (du/x⁵) = (1/6) ∫ du/(x³u^(2/3))</p><p><strong>Step 4:</strong> From u = 1 + x⁶, differentiate the proposed answer: d/dx[xf(x)(1+x⁶)^(1/3)] = f(x)(1+x⁶)^(1/3) + xf'(x)(1+x⁶)^(1/3) + xf(x)·(1/3)(1+x⁶)^(-2/3)·6x⁵</p><p><strong>Step 5:</strong> Matching coefficients with 1/[x³(1+x⁶)^(2/3)], we get: f(x) = -1/(6x²)</p><p>∴ Answer: f(x) = <strong>-1/(6x²)</strong></p>
Correct Answer: B