Indefinite Integration
Integration by parts
Grade 12
Question:
<p><strong>331.</strong> If \(\displaystyle\int x \ln\left(1 + \dfrac{1}{x}\right)dx = f(x)\ln(x+1) + g(x)x^2 + kx + C\), where \(C\) is constant of integration, then:</p>
<p>\(\lim_{x \to 0} \dfrac{f(\cos x)}{x^2} = \dfrac{-1}{2}\)</p>
<p>\(\lim_{x \to 0} \dfrac{g(1+x)}{x} = \dfrac{-1}{2}\)</p>
<p>\(\lim_{x \to 0}(1 + f(x) + k)^{\frac{1}{x - \sin x}} = e^{-3}\)</p>
<p>\(\displaystyle\int_1^e \dfrac{g(x)}{k}\,dx = 1\)</p>
Step-by-Step Solution
Key Concept: Simplify the integrand using algebraic manipulation: x·ln(1+1/x) = x·ln((x+1)/x) = x[ln(x+1) - ln(x)], then integrate by parts strategically to match the given form.
<p><strong>Step 1:</strong> Simplify the integrand</p><p>x·ln(1 + 1/x) = x·ln((x+1)/x) = x[ln(x+1) - ln(x)]</p><p><strong>Step 2:</strong> Expand and separate</p><p>∫x·ln(1 + 1/x)dx = ∫x·ln(x+1)dx - ∫x·ln(x)dx</p><p><strong>Step 3:</strong> Apply integration by parts to ∫x·ln(x+1)dx</p><p>Let u = ln(x+1), dv = x dx → du = 1/(x+1)dx, v = x²/2</p><p>∫x·ln(x+1)dx = (x²/2)ln(x+1) - ∫(x²/2)·1/(x+1)dx</p><p><strong>Step 4:</strong> Simplify x²/(2(x+1)) using polynomial division</p><p>x²/(2(x+1)) = x/2 - 1/2 + 1/(2(x+1))</p><p>∫(x²/2(x+1))dx = x²/4 - x/2 + (1/2)ln(x+1)</p><p><strong>Step 5:</strong> Apply integration by parts to ∫x·ln(x)dx</p><p>Similarly: ∫x·ln(x)dx = (x²/2)ln(x) - x²/4</p><p><strong>Step 6:</strong> Combine results</p><p>∫x·ln(1 + 1/x)dx = (x²/2)ln(x+1) - x²/4 + x/2 - (1/2)ln(x+1) - [(x²/2)ln(x) - x²/4]</p><p>= [(x²/2) - 1/2]ln(x+1) - (x²/2)ln(x) + x/2 + C</p><p><strong>Step 7:</strong> Identify the form</p><p>f(x) = x²/2 - 1/2, g(x) = -1/2, k = 1/2</p><p>∴ Answer: Compare coefficients and verify the match with given form</p>
Correct Answer: A,B,C,D