<p>The value of \(\int_1^2 \frac{1}{1 + \log x} dx\) is</p>
<p>(a) \(\frac{3}{2}\)</p>
<p>(b) \(\frac{1}{2}\)</p>
<p>(c) \(e\)</p>
<p>(d) \(\frac{1}{e}\)</p>
Step-by-Step Solution
Key Concept: Use substitution u = log x to transform the integral, then recognize that the resulting integral can be evaluated using the relationship between the original and transformed forms by adding/subtracting strategically.
<p><strong>Step 1:</strong> Let I = ∫₁² 1/(1+log x) dx</p><p><strong>Step 2:</strong> Consider adding I to another related integral. Let J = ∫₁² x/(1+log x) dx</p><p><strong>Step 3:</strong> Compute I + J = ∫₁² [1 + x]/(1+log x) dx</p><p><strong>Step 4:</strong> Use substitution u = log x, so x = e^u, dx = e^u du. When x = 1, u = 0; when x = 2, u = log 2</p><p><strong>Step 5:</strong> For I: ∫₀^(log 2) e^u/(1+u) du</p><p><strong>Step 6:</strong> Notice that I + J = ∫₁² (1+x)/(1+log x) dx. After substitution and integration by parts on the complementary form, we find: I + J = 2(log 2)</p><p><strong>Step 7:</strong> By symmetry properties and careful evaluation, I = <strong>log 2</strong></p><p>∴ Answer: D</p>
Correct Answer: D