\[I= \int^1_0 \frac{x-1}{lnx} dx\]
consider \[F( \alpha )= \int^1_0 \frac{x^{\alpha }}{lnx} dx\]
/\[I=F(1)\]
.\[\frac{dF(\alpha)}{d \alpha}= \int^1_0 \frac{x^{\alpha} lnx}{lnx} dx= \int^1_0 x^{\alpha} dx= \frac{1}{\alpha +1}\]
.Then
\[F (\alpha )= \int \frac{1}{\alpha +1} d \alpha = ln( \alpha +1) +c\]
.When
\[\alpha =0\]
\[F(0 )= \int^1_0 \frac{1-1}{lnx} dx=0\]
so \[0=0+c \rightarrow c=0\]
.Then
\[I= \int^1_0 \frac{x-1}{lnx} dx = lim_{\alpha \rightarrow 1 }\int^1_0 \frac{x^{\alpha }-1}{lnx} dx=F(1)=ln2\]
.