Answer
$[-2, -1) ∪ (0, 1]$
Work Step by Step
$$1+\frac{2}{x+1}\leq \frac{2}{x}$$
Subtract $\frac{2}{x}$ from both sides
$$1+\frac{2}{x+1}-\frac{2}{x}\leq0$$
And simplify
$$\frac{x(x+1)+2x-2(x+1)}{x(x+1)}\leq0$$
$$\frac{x^2+x+2x-2x-2}{x(x+1)}\leq0$$
$$\frac{x^2+x-2}{x(x+1)}\leq0$$
Factor out in numerator
$$\frac{(x-1)(x+2)}{x(x+1)}\leq0$$
We have key points at:
$x-1=0$ ; $x=1$
$x+2=0$ ; $x=-2$
$x=0$
$x+1=0$ ; $x=-1$
Which gives us intervals:
$(-\infty, -2]$ - Positive
$[-2, -1)$ - Negative
$(-1, 0)$ - Positive
$(0, 1]$ - Negative
$[1, \infty)$ - Positive
We need closed intervals of negative numbers, so the answer is:
$[-2, -1) ∪ (0, 1]$
