How many integers satisfy the inequality?

We can remove the absolute value sign by squaring both sides, and solving

$\left( \frac {10(x+1) } {x^2 + 2x + 3} \right)^2 \geq 1^2$

We can multiply by $(x^2 + 2x + 3)^2$ which is always positive, to get

$[ 10(x+1) ] ^2 \geq (x^2 + 2x + 3)^2$

Shifting terms to one side and using the factorization $a^2 - b^2 = (a-b)(a+b)$ , we get

$0 \geq (x^2 + 12x + 13 ) ( x^2 -8x - 7)$

The expression on the right crosses the x-axis at the zeros, which are $\frac {-12 - \sqrt{92} } {2}, \frac {-12 + \sqrt{92}}{2} , \frac {8 - \sqrt{92} }{2}, \frac { 8 + \sqrt{92} } {2}$ .

Hence, the inequality is satisfied when $x = -10, -9, -8, -7, -6, -5, -4, -3, -2, 0, 1, 2, 3, 4, 5, 6, 7, 8$ , for a total of 18 values.