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

1. Problem Description

The problem is 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 $x^2+2$ is always positive for 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:

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

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

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

$2x^2 - x < 0$

$x(2x-1) < 0$

The roots of $x(2x-1)=0$ are $x=0$ and $x=\frac{1}{2}$. The parabola $y=x(2x-1)$ opens upwards, so the inequality $x(2x-1)<0$ is satisfied when $0 < x < \frac{1}{2}$.

However, we assumed that $x \ge 2$, so 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 assumed $x < 2$, and we found $x < -4$, the solution for this case is $x < -4$.

3. Final Answer

$x < -4$