$Lnx=u \Rightarrow x=e^u \Rightarrow dx=e^udu$
$\Rightarrow \int _0^1 \frac{x-1-Lnx}{xLnx-xLn^2x}= \int_{- \infty }^0 \frac{e^u (e^u-1-u)}{ue^u-u^2e^u}du$
$=\int _{- \infty }^0\frac{e^u-1-u}{u-u^2} du= \int _{- \infty }^0 \frac{e^u}{u(1-u)} du- \int _{- \infty }^0 \frac{1+u}{u(1-u)} du$
$= \int _{- \infty }^0 \frac{e^u}{u} du+\int _{- \infty }^0 \frac{e^u}{u-1} du-\int _{- \infty }^0( \frac{1}{u}+ \frac{2}{1-u} ) du$
$=\int _{- \infty }^0 \frac{e^u}{u} du+e\int _{- \infty }^{-1} \frac{e^v}{v} dv-\int _{- \infty }^0 (\frac{1}{u}+ \frac{2}{u-1} )du$
$=Ei(0)+Ei(-1)- \int_{- \infty }^0 ( \frac{1}{u} +\frac{2}{1-u} )du$
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