<p>Evaluate \(I = \int \dfrac{1}{2e^{2x} + 3e^x + 1}\,dx\). The answer involves a logarithmic expression. Find the integer value associated with the result.</p>
Step-by-Step Solution
Key Concept: Factor the denominator as a quadratic in e^x: 2e^(2x) + 3e^x + 1 = (2e^x + 1)(e^x + 1). Use partial fraction decomposition to split the integrand into simpler terms that integrate to logarithms.
<p><strong>Step 1:</strong> Recognize the denominator as a quadratic in e^x. Let u = e^x, so the denominator becomes 2u² + 3u + 1.</p><p><strong>Step 2:</strong> Factor: 2u² + 3u + 1 = (2u + 1)(u + 1) = (2e^x + 1)(e^x + 1)</p><p><strong>Step 3:</strong> Apply partial fractions: <br/>$$\frac{1}{(2e^x + 1)(e^x + 1)} = \frac{A}{2e^x + 1} + \frac{B}{e^x + 1}$$</p><p><strong>Step 4:</strong> Solving for A and B: 1 = A(e^x + 1) + B(2e^x + 1)<br/>Setting e^x = -1/2: 1 = A(1/2) → A = 2<br/>Setting e^x = -1: 1 = B(-1) → B = -1</p><p><strong>Step 5:</strong> Integrate:<br/>$$I = \int \frac{2}{2e^x + 1}\,dx - \int \frac{1}{e^x + 1}\,dx$$<br/>$$= \ln|2e^x + 1| - \ln|e^x + 1| + C$$<br/>$$= \ln\left|\frac{2e^x + 1}{e^x + 1}\right| + C$$</p><p><strong>Step 6:</strong> The coefficient of the logarithmic expression is 1 (from the ratio argument), and the integer associated with the answer structure is <strong>2</strong> (from the numerator factor 2e^x in the ratio).</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2