The problem is to solve the inequality $\frac{(x+2)|x-3|}{x^2+6} > -1$.

1. Problem Description

The problem is to solve the inequality $\frac{(x+2)|x-3|}{x^2+6} > -1$.

2. Solution Steps

First, we multiply both sides of the inequality by $x^2+6$. Since $x^2+6$ is always positive for all real numbers $x$, the inequality sign does not change.

$(x+2)|x-3| > -(x^2+6)$

$(x+2)|x-3| > -x^2-6$

$(x+2)|x-3| + x^2 + 6 > 0$

Consider the case where $x \geq 3$, so $|x-3| = x-3$.

$(x+2)(x-3) + x^2 + 6 > 0$

$x^2 -3x + 2x -6 + x^2 + 6 > 0$

$2x^2 -x > 0$

$x(2x-1) > 0$

This inequality holds if $x > 0$ and $2x-1 > 0$ or $x < 0$ and $2x-1 < 0$.

$x > 0$ and $x > \frac{1}{2}$ gives $x > \frac{1}{2}$.

$x < 0$ and $x < \frac{1}{2}$ gives $x < 0$.

However, we are considering the case where $x \geq 3$. So, we consider the intersection of $x \geq 3$ with $(x < 0$ or $x > \frac{1}{2})$. This gives $x \geq 3$.

Consider the case where $x < 3$, so $|x-3| = -(x-3) = 3-x$.

$(x+2)(3-x) + x^2 + 6 > 0$

$3x -x^2 + 6 -2x + x^2 + 6 > 0$

$x + 12 > 0$

$x > -12$

However, we are considering the case where $x < 3$. So we consider the intersection of $x < 3$ with $x > -12$. This gives $-12 < x < 3$.

Combining both cases, we have $x \geq 3$ or $-12 < x < 3$. This gives us $x > -12$.

Final Answer: The final answer is $\boxed{x>-12}$