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

1. Problem Description

We need 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 $x^2+6 > 0$ for all real $x$, the inequality sign does not change.

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

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

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

We consider two cases:

Case 1: $x \geq 3$. In this case, $|x-3| = x-3$, so the inequality becomes:

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

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

$2x^2 - x > 0$

$x(2x-1) > 0$

This implies either $x > 0$ and $2x-1 > 0$, or $x < 0$ and $2x-1 < 0$.

$x > 0$ and $x > \frac{1}{2}$, which means $x > \frac{1}{2}$.

$x < 0$ and $x < \frac{1}{2}$, which means $x < 0$.

So $x < 0$ or $x > \frac{1}{2}$.

Since we assumed $x \geq 3$, the solution in this case is $x \geq 3$.

Case 2: $x < 3$. In this case, $|x-3| = -(x-3) = 3-x$, so the inequality becomes:

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

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

$x + 12 > 0$

$x > -12$

Since we assumed $x < 3$, the solution in this case is $-12 < x < 3$.

Combining both cases:

$x \geq 3$ or $-12 < x < 3$, which means $x > -12$.

Therefore, the solution to the inequality is $x > -12$.

3. Final Answer

$x > -12$