Evaluating $\int_0^\infty \frac{\ln |x-1|}{x^2((\ln x)^2-(\ln x)(\ln|x-1|))} dx$ 
The question is to evaluate $$\int_0^\infty \frac{\ln |x-1|}{x^2((\ln x)^2-(\ln x)(\ln|x-1|))} dx$$

I tried using the substitution $t=1/x$ which rewrites the integral to $$-\int_\infty^0 \frac{\ln \frac{|1-x|}{x}}{(\ln x)^2+(\ln x)\ln\frac{|1-x|}{x}} dx$$ which can again be rewritten as $$\int^\infty_0 \frac{\ln \frac{|1-x|}{x}}{\ln x\ln |1-x|} dx$$ which is same as  $$\int^\infty_0 \frac{1}{\ln x}-\frac{1}{\ln |1-x|}dx$$ I could not proceed after this.Any ideas?Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The integral $\color{#f00}{diverges}$ because
\begin{align}
&{\ln\pars{\verts{x - 1}} \over x^{2}
\bracks{\vphantom{\Large A}%
\ln^{2}\pars{x} - \ln\pars{x}\ln\pars{\verts{x - 1}}}}
\stackrel{\mrm{as}\ x\ \to\ \infty}{\sim}
{\ln\pars{x} \over x^{2}\braces{\vphantom{\Large A}%
\ln^{2}\pars{x} - \ln\pars{x}\bracks{\vphantom{\large A}\ln\pars{x} - 1/x}}} = \color{#f00}{1 \over x}
\end{align}
