0\; \; pri \; \; x\in (-\infty .+\infty )\; ,\; t.k.\; \; D=1-4=-3<0\; \to \\\\(x-1)(x+1)<0\quad +++(-1)---(1)+++\\\\x\in (-1,1)\\\\celue\; x:\; \; x=0\in (-1,1)" alt=" (\frac{x}{x-1})^2-\frac{1}{x+1}\leq\frac{2x}{(x-1)^2(x+1)}\\\\\frac{x^2}{(x-1)^2}-\frac{1}{x+1}-\frac{2x}{(x-1)^2(x+1)}\leq 0\\\\\frac{x^2(x+1)-(x-1)^2-2x}{(x-1)^2(x+1)}\leq 0\\\\\frac{x^3+x^2-(x^2-2x+1)-2x}{(x-1)^2(x+1)}\leq 0\\\\\frac{x^3-1}{(x-1)^2(x+1)}\leq 0\\\\\frac{(x-1)(x^2+x+1)}{(x-1)^2(x+1)}\leq 0\\\\\frac{x^2+x+1}{(x-1)(x+1)}\leq 0\\\\x^2+x+1>0\; \; pri \; \; x\in (-\infty .+\infty )\; ,\; t.k.\; \; D=1-4=-3<0\; \to \\\\(x-1)(x+1)<0\quad +++(-1)---(1)+++\\\\x\in (-1,1)\\\\celue\; x:\; \; x=0\in (-1,1)" align="absmiddle" class="latex-formula">
Ответ: количество целых решений неравенства - одно (ответ №2).