\(\dfrac{{x + 1}}{{x-1}} > x + 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{x + 1}}{{x-1}}-\dfrac{{x + 1}}{1} > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{x + 1-\left( {x + 1} \right)\left( {x-1} \right)}}{{x-1}} > 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\left( {x + 1} \right)\left( {1-x + 1} \right)}}{{x-1}} > 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\left( {x + 1} \right)\left( {x-2} \right)}}{{x-1}} < 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\(\left( {x + 1} \right)\left( {x-2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,}\\{x-2 = 0\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 2.\,\,}\end{array}} \right.\)
Найдём нули знаменателя: \(x-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 1.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;2} \right).\)
Ответ: \(\left( {-\infty ;\;-1} \right) \cup \left( {1;\;2} \right).\)