The correct option is D None of these
∫∞1log(t−1)t2logtlogt(t−1)dt=∫10log(1x−1)logxlog(1−x)dx
(By substituting t=1xdt=1x2dxandt=1⇔x=1andt=∞⇔x=0)
∴I=∫10log(1−x)−logxlogxlog(1−x)dx
=∫10(1logx−1log(1−x))dx
=∫10(1log(1−x)−1logx)dx=−I∴I=0
[∵∫a0f(x)dx=∫a0f(a−x)dx]