Number of integral values of x satisfying the inequation  $\frac{x}{x+2}\le \frac{1}{\left|x\right|}\text{ is }$

$\frac{x\left|x\right|}{x+2}\le 1$

$\frac{x\left|x\right|-x-2}{x+2}\le 0$

$$\begin{gathered} \begin{array}{l}\text{ Case I: }x\in [0,\mathrm{\infty })\\ \begin{array}{c}\frac{x^2-x-2}{x+2}\le 0\\ \Rightarrow \frac{(x-2)(x+1)}{x+2}\le 0\end{array}\end{array} \\ \Rightarrow 0\le x\le 2 \left(\because x\in [0,\mathrm{\infty })\right) \end{gathered}$$

i.e., integral values of x are 0, 1 , 2.

$\begin{array}{r}\text{ Case II: }x\in (-\mathrm{\infty },0)\\ \frac{-x^2-x-2}{x+2}\le 0\end{array}$

$\begin{array}{l}\Rightarrow -2<x<0\\ \Rightarrow x=-1\end{array}(\because x\in (-\mathrm{\infty },0\left)\right)$