The value of \(\int_{0}^{1}\displaystyle [\log x - \log(1 - x)]\, dx\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
Let I be the integral. By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) with a = 1, x is replaced by 1 − x. The two logarithms change places:
\( \displaystyle I = \int_{0}^{1} [\log(1 - x) - \log x]\,dx \)
\( \displaystyle I = -\int_{0}^{1} [\log x - \log(1 - x)]\,dx = -I \)
\(2I = 0,\quad I = 0\)