Question

\( \int_{0}^{1} \log (\frac{1}{x} - 1) dx = \)}

• A. \( \frac{1}{2} \)
• B. 1
• C. 2
• D. 0

Hint:

The property $\int_0^a f(x) = \int_0^a f(a-x)$ is extremely powerful for log-rational functions.

Answer 1

The Correct Option is D

Step 1: Use Integral Property
$\int_0^a f(x) dx = \int_0^a f(a-x) dx$. $I = \int_0^1 \log(\frac{1-x}{x}) dx$.
Step 2: Apply Property
$I = \int_0^1 \log(\frac{1-(1-x)}{1-x}) dx = \int_0^1 \log(\frac{x}{1-x}) dx$.
Step 3: Add Integrals
$2I = \int_0^1 [\log(\frac{1-x}{x}) + \log(\frac{x}{1-x})] dx$. $2I = \int_0^1 \log(\frac{1-x}{x} \cdot \frac{x}{1-x}) dx = \int_0^1 \log(1) dx = 0$.
Step 4: Conclusion
$I = 0$.
Final Answer:(D)