We need to solve the inequality $\frac{(x+1)|x-2|}{x^2+2} < -1$.

1. Problem Description

We need to solve the inequality $\frac{(x+1)|x-2|}{x^2+2} < -1$.

2. Solution Steps

First, we multiply both sides of the inequality by $x^2+2$. Since $x^2+2 > 0$ for all real $x$, the inequality sign remains the same.

$(x+1)|x-2| < -(x^2+2)$

$(x+1)|x-2| < -x^2-2$

$(x+1)|x-2| + x^2 + 2 < 0$

We consider two cases: $x \ge 2$ and $x < 2$.

Case 1: $x \ge 2$. In this case, $|x-2| = x-2$.

$(x+1)(x-2) + x^2 + 2 < 0$

$x^2 - 2x + x - 2 + x^2 + 2 < 0$

$2x^2 - x < 0$

$x(2x - 1) < 0$

$0 < x < \frac{1}{2}$

However, we have the condition $x \ge 2$. Since there is no overlap between $0 < x < \frac{1}{2}$ and $x \ge 2$, there are no solutions in this case.

Case 2: $x < 2$. In this case, $|x-2| = -(x-2) = 2-x$.

$(x+1)(2-x) + x^2 + 2 < 0$

$2x - x^2 + 2 - x + x^2 + 2 < 0$

$x + 4 < 0$

$x < -4$

Since we have the condition $x < 2$, the intersection of $x < -4$ and $x < 2$ is $x < -4$.

3. Final Answer

The solution is $x < -4$.