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The value of $\int\limits_0^1 [\log x - \log(1 - x)] dx$ is |
$\frac{1}{2}$ 2 1 0 |
0 |
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 $I=-I \Rightarrow 2I=0 \Rightarrow I=0$ |
