$\text{ Solution of the inequation }{\log }_{0.1}\left({\log }_2\frac{x^2+1}{|x-1|}\right)<0\text{ contains the interval }$

$\begin{array}{l}{\log }_2\left(\frac{x^2+1}{|x-1|}\right)>1\Rightarrow \frac{x^2+1}{|x-1|}>2\\ \text{ If }x>1x^2+1>2|x-1|\\ \text{ or }{x}^2-2x+3>0,x\in R\\ \text{ If }x<1x^2+1>2-2x\\ \text{ or }{x}^2+2x-1>0\end{array}$

$\begin{array}{l}\Rightarrow (x-(-1-\sqrt{2}\left)\right(x-(-1+\sqrt{2})>0\\ \text{ thus, }x\in (-\mathrm{\infty },-1-\sqrt{2})\cup (\sqrt{2}-1,1)\cup (1,\mathrm{\infty })\end{array}$