<p><strong>311.</strong> Let \(\displaystyle\int \dfrac{(x-1)e^x}{(x+1)^3}\,dx = f(x) + C\) where \(f(x) = d + \displaystyle\sum_{i=0}^{n} \dfrac{a_i e^x}{(x+1)^i}\) with all \(a_i = 0\) for \(i \geq n\) and \(f(1) = \dfrac{e}{2}\). Then which of the following is/are correct?</p>
Step-by-Step Solution
Key Concept: Decompose the integrand using partial fractions with an exponential factor, then recognize that repeated integration by parts or strategic algebraic manipulation reduces the denominator power. The condition f(1) = e/2 determines the constant d and validates the form.
<p><strong>Step 1:</strong> Rewrite the numerator strategically. Notice that (x-1) = (x+1) - 2, so:</p><p>∫[(x-1)e^x/(x+1)³]dx = ∫[e^x/(x+1)²]dx - 2∫[e^x/(x+1)³]dx</p><p><strong>Step 2:</strong> For ∫[e^x/(x+1)²]dx, use integration by parts or recognize that d/dx[e^x/(x+1)] = e^x/(x+1) - e^x/(x+1)². This gives:</p><p>∫[e^x/(x+1)²]dx = e^x/(x+1) + ∫[e^x/(x+1)²]dx (iterate)</p><p><strong>Step 3:</strong> Alternative: Note that if f(x) = d + a₀e^x + a₁e^x/(x+1) + a₂e^x/(x+1)², then f'(x) must equal (x-1)e^x/(x+1)³.</p><p>Differentiating: f'(x) = a₀e^x + [a₁e^x(x+1) - a₁e^x]/(x+1)² + [a₂e^x(x+1)² - 2a₂e^x(x+1)]/(x+1)³</p><p>= a₀e^x + a₁e^x/(x+1) + a₂e^x[(x+1) - 2]/(x+1)³</p><p>= a₀e^x + a₁e^x/(x+1) + a₂e^x(x-1)/(x+1)³</p><p><strong>Step 4:</strong> Matching with (x-1)e^x/(x+1)³: Set a₂ = 1, a₁ = 0, a₀ = 0.</p><p><strong>Step 5:</strong> So f(x) = d + e^x/(x+1)². Apply f(1) = e/2:</p><p>d + e/(2)² = e/2 ⟹ d + e/4 = e/2 ⟹ d = e/4</p><p>∴ f(x) = e/4 + e^x/(x+1)², with n = 2, a₀ = 0, a₁ = 0, a₂ = 1</p>
Correct Answer: ACD