Let I = \(\int \limits_0^1 \log(\frac 1x - 1)dx\)
\(= \int \limits_0^1 \log(\frac {1 -x}x)dx\) ......(i)
\(I = \int \limits_{0}^1 \log\left(\frac{1-(1 -x)}{1-x}\right)dx\)
\(I = \int \limits_0^1 \log\left(\frac x{1-x}\right) dx\) ......(ii)
Adding (i) and (ii), we get
\(2I = \int \limits_0^1 \log\left(\frac {1-x}x\right) dx + \int \limits_0^1 \log\left(\frac x{1-x}\right) dx\)
\(= \int \limits_0^1 \log\left(\frac{1 -x}x.\frac x {1 -x}\right)dx\)
\(= \int \limits_{0}^1 \log 1 dx\)
\(= 0\)
\(\therefore I = 0\)
