The value of $\int\limits_0^1 [\log x - \log(1 - x)] dx$ is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
Given: $I=\int_{0}^{1}[\log x - \log(1-x)]\,dx$
Substitute $t=1-x \Rightarrow dx=-dt$:
$I=\int_{1}^{0}[\log(1-t)-\log t](-dt)=\int_{0}^{1}[\log(1-t)-\log t]\,dt$
Thus $I=-I \Rightarrow 2I=0 \Rightarrow I=0$