The given inequality is:$$x^5+x^4-6x^3+x^2+x-6\ge0.$$Let's solve it by factoring the expression and finding the solution to the inequality. First, we can factor the given polynomial as:$$x^5+x^4-6x^3+x^2+x-6=(x-1)(x+2)(x^3-3x^2+x+3).$$Therefore, the inequality can be rewritten as$$(x-1)(x+2)(x^3-3x^2+x+3)\ge 0.$$Now, we can solve this inequality by analyzing the sign of each factor in the three intervals where the entire real line is divided:$$\begin{array}ccccccccccc x & -\infty & & -2 & & -1 & & 1 & & & 2 & & \infty \\ \hline (x-1) & - & - & - & - & - & 0 & + & + & + & + & + & + \\ (x+2) & - & - & - & 0 & + & + & + & + & + & + & + & + \\ (x^3-3x^2+x+3) & - & - & + & + & + & + & + & + & + & + & + & + \\ \hline (x-1)(x+2)(x^3-3x^2+x+3) & - & + & - & 0 & - & 0 & + & + & + & + & + & + \\ \end{array}$$Thus, the solution set of the inequality is $(-\infty,-2]\cup[-1,2]\cup[2,\infty)$, which is option D.