jmc

From the given inequality, either $\frac{3(x+1)}{x^2+2x+3}\ge 1$ or $\frac{3(x+1)}{x^2+2x+3}\le -1.$

We start with the inequality $\frac{3(x+1)}{x^2+2x+3}\ge 1.$ Since $x^2+2x+3=(x+1{)}^2+2$ is always positive, we can safely multiply both sides by $x^2+2x+3,$ to get $$3x+3\ge x^2+2x+3.$$Then $x^2-x\le 0,$ or $x(x-1)\le 0.$ This is satisfied for $0\le x\le 1.$

Next, we look at the inequality $\frac{3(x+1)}{x^2+2x+3}\le -1.$ Again, we can safely multiply both sides by $x^2+2x+3,$ to get $$3x+3\le -x^2-2x-3.$$Then $x^2+5x+6\le 0,$ or $(x+2)(x+3)\le 0.$ This is satisfied for $-3\le x\le -2.$

Thus, the solution is $x\in \boxed{[-3,-2]\cup [0,1]}.$