Solve: `(log)_(0. 1)((log)_2((x^2+1)/(x-1))<0`

Explanation

` log_(0.1)(log_(2).(x^(2)+1)/(|x-1|))lt 0`
` or log_(2)((x^(2)+1)/(|x-1|)) gt 1`
` or (x^(2)+1)/(|x-1|) gt 2`
Case (i): `x gt 1`
`(x^(2)+1)/(x-1) gt 2`
` or x^(2)+1 gt 2x-2`
` or x^(2) - 2x + 3 gt 0`,which is always true
` rArr x gt 1`
Case(ii): ` x lt 1`
` (x^(2)+1)/(1-x) gt 2`
` or x^(2) + 1 gt 2 - 2x`
` or x^(2)+ 2x - 1 gt 0`
` rArr x lt - 1 - sqrt2 or x gt 1+ sqrt2`
Hence `x in (-infty, -1 - sqrt2) cup(sqrt2 - 1, 1) cup (1, infty)`.