Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>If <em>m</em> is a non-zero number and \(\int \frac{x^{5m-1}+2x^{4m-1}}{(x^{2m}+x^m+1)^3}dx = f(x)+C\), then <em>f(x)</em> is</p>
<p>\(\frac{x^{5m}}{2m(x^{2m}+x^m+1)^2}\)</p>
<p>\(\frac{x^{4m}}{2m(x^{2m}+x^m+1)^2}\)</p>
<p>\(\frac{2m(x^{5m}+x^{4m})}{(x^{2m}+x^m+1)^2}\)</p>
<p>\(\frac{(x^{5m}-x^{4m})}{2m(x^{2m}+x^m+1)^2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator is the derivative of the denominator's base expression multiplied by a constant. Use substitution u = x^m to reduce the integral to a rational function, then apply the chain rule backwards.
<p><strong>Step 1:</strong> Let u = x^m, then du = mx^(m-1)dx, so x^(m-1)dx = du/m</p><p><strong>Step 2:</strong> Rewrite the numerator: x^(5m-1) + 2x^(4m-1) = x^(m-1)·x^(4m) + 2x^(m-1)·x^(3m) = x^(m-1)(x^(4m) + 2x^(3m))</p><p><strong>Step 3:</strong> Notice that x^(4m) + 2x^(3m) = u^4 + 2u^3 and the integral becomes:</p><p>∫ (u^4 + 2u^3)/(x^m + x^m + 1)^3 · (du/m) = (1/m)∫ (u^4 + 2u^3)/(u^2 + u + 1)^3 du</p><p><strong>Step 4:</strong> Recognize that d/du[u^2 + u + 1] = 2u + 1, so (u^4 + 2u^3) = u^3(u + 2). Using the pattern, observe:</p><p>(1/m)∫ d/du[-u^2/(2(u^2+u+1)^2)] du</p><p><strong>Step 5:</strong> Therefore: f(x) = <strong>-x^(2m)/(2m(x^(2m) + x^m + 1)^2)</strong></p><p>∴ Answer: A</p>
Correct Answer: A