Indefinite Integration
Integration by substitution
Grade None
Question:
<p>\(\int \frac{dx}{x(x^n+1)}\) is equal to</p>
<p>\(\frac{1}{n}\log\left(\frac{x^n}{x^n+1}\right)+C\)</p>
<p>\(\frac{1}{n}\log\left(\frac{x^n+1}{x^n}\right)+C\)</p>
<p>\(\log\left(\frac{x^n}{x^n+1}\right)+C\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use partial fractions with the substitution technique by recognizing that d(x^n)/dx = nx^(n-1), allowing us to decompose 1/[x(x^n+1)] into manageable terms involving x^n.
<p><strong>Step 1:</strong> Multiply numerator and denominator by x^(n-1):</p><p>∫ dx/[x(x^n+1)] = ∫ x^(n-1)·dx/[x^n(x^n+1)]</p><p><strong>Step 2:</strong> Use partial fractions on x^(n-1)/[x^n(x^n+1)]. Rewrite: x^(n-1) = (1/n)·nx^(n-1) = (1/n)[d(x^n)/dx]</p><p><strong>Step 3:</strong> Decompose: 1/[x^n(x^n+1)] = 1/x^n - 1/(x^n+1)</p><p><strong>Step 4:</strong> Therefore:</p><p>∫ dx/[x(x^n+1)] = (1/n)∫[1/x^n - 1/(x^n+1)]·d(x^n)</p><p><strong>Step 5:</strong> Let u = x^n, then:</p><p>= (1/n)∫[1/u - 1/(u+1)]du = (1/n)[ln|u| - ln|u+1|] + C</p><p>= (1/n)ln|x^n/(x^n+1)| + C</p><p>∴ Answer: <strong>(1/n)ln|x^n/(x^n+1)| + C</strong></p>
Correct Answer: A