Rewrite the expression
5logx4log(x2+1)+4log(x1)\displaystyle {5}{\log{{x}}}-{4}{\log{{\left({x}^{{2}}+{1}\right)}}}+{4}{\log{{\left({x}-{1}\right)}}}
as a single logarithm logA\displaystyle {\log{{A}}}. Then the function
A=\displaystyle {A}=