<p>If \( f(x) = \displaystyle\int \frac{3x^2 - x^{-2}}{\left(x^3 + 1 + \dfrac{1}{x}\right)^2}\, dx \) and \( f(0) = 0 \), then \( f(-1) \) equals:</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator is the derivative of the denominator's base expression (x³ + 1 + 1/x). Use substitution u = x³ + 1 + 1/x to convert this into a standard form ∫du/u².
<p><strong>Step 1:</strong> Recognize the substitution pattern. Let u = x³ + 1 + 1/x</p><p><strong>Step 2:</strong> Compute du/dx = 3x² - x⁻² = 3x² - 1/x². This exactly matches our numerator!</p><p><strong>Step 3:</strong> Rewrite the integral: f(x) = ∫(du/dx)/u² dx = ∫u⁻² du = -u⁻¹ + C = -1/(x³ + 1 + 1/x) + C</p><p><strong>Step 4:</strong> Apply initial condition f(0) = 0. As x → 0, the denominator (x³ + 1 + 1/x) → ∞, but we need to be careful. Note: x = 0 is not in the domain, so we interpret f(0) = 0 as the constant C. Actually, rewrite: f(x) = -1/(x³ + 1 + 1/x) + C</p><p><strong>Step 5:</strong> For the boundary condition to make sense with the domain, we determine C such that the function matches given conditions. With standard convention: C = 0.</p><p><strong>Step 6:</strong> Calculate f(-1): f(-1) = -1/((-1)³ + 1 + 1/(-1)) = -1/(-1 + 1 - 1) = -1/(-1) = 1</p><p>∴ Answer: C (f(-1) = 1)</p>
Correct Answer: C