The value of \(\int_{0}^{1}\displaystyle \log_e \left(\frac{1}{x}-1\right) dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
Let I be the integral. Writing the bracket as one fraction:
\( \displaystyle I = \int_{0}^{1} \log_e\left(\frac{1-x}{x}\right) dx \)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) (King's property), replace x by 1 − x:
\( \displaystyle I = \int_{0}^{1} \log_e\left(\frac{x}{1-x}\right) dx \)
\( \displaystyle = -\int_{0}^{1} \log_e\left(\frac{1-x}{x}\right) dx = -I \)
\(2I = 0\)
\(I = 0\)