The value of $\int\limits_0^1\log_e (\frac{1}{x}-1) dx$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
Evaluate the integral:
$\int_{0}^{1} \log\!\left(\frac{1}{x}-1\right)\,dx$
Simplify the expression:
$\frac{1}{x}-1=\frac{1-x}{x}$
$\log\!\left(\frac{1-x}{x}\right)=\log(1-x)-\log(x)$
Thus:
$\int_{0}^{1}\log(1-x)\,dx-\int_{0}^{1}\log(x)\,dx$
Use known results:
$\int_{0}^{1}\log x\,dx=-1$
$\int_{0}^{1}\log(1-x)\,dx=-1$
Therefore:
$-1 - (-1)=0$
The value of the integral is 0.