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: Understanding the Question:
Apply the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx \).
Step 2: Detailed Explanation:
Let \( I = \int_{0}^{1} \log(\frac{1-x}{x}) dx \dots (i) \)
Using the property:
\[ I = \int_{0}^{1} \log(\frac{1-(1-x)}{1-x}) dx = \int_{0}^{1} \log(\frac{x}{1-x}) dx \dots (ii) \] Add (i) and (ii):
\[ 2I = \int_{0}^{1} [\log(\frac{1-x}{x}) + \log(\frac{x}{1-x})] dx \] Using log property \( \log A + \log B = \log(AB) \):
\[ 2I = \int_{0}^{1} \log(\frac{1-x}{x} \cdot \frac{x}{1-x}) dx = \int_{0}^{1} \log(1) dx = \int_{0}^{1} 0 dx = 0 \] So \( I = 0 \).
Step 4: Final Answer:
The integral is 0.