The possible values of x, which satisfy the trigonometric equation

$${\tan ^{ - 1}}\left( {{{x - 1} \over {x - 2}}} \right) + {\tan ^{ - 1}}\left( {{{x + 1} \over {x + 2}}} \right) = {\pi \over 4}$$ are

If $$f(x) = {\tan ^{ - 1}}\left[ {{{\log \left( {{e \over {{x^2}}}} \right)} \over {\log (e{x^2})}}} \right] + {\tan ^{ - 1}}\left[ {{{3 + 2\log x} \over {1 - 6\log x}}} \right]$$, then the value of f''(x) is equal to