I am not sure if the answer for the dividing rational expressions problem should be simplified $$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$
-I am not sure if the answer for the problem (which is attached) would be $\frac{x^3-4x}{x+2}$ or if it could be simplified further.

Remember that:
$$a \div \dfrac{1}{c}=ac$$
Therefore we can convert
$$\frac{x}{x+2} \div \dfrac{1}{x^2-4}$$
into:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)$$
We can simplify this by factoring $x^2-4$ using the difference of squares formula, which is:
$$a^2-b^2=(a+b)(a-b)$$
So:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)=\left(\frac{x}{x+2}\right)(x+2)(x-2)$$
Do you see that we can cancel $x+2$ out?
$$\require{cancel}{\left(\frac{x}{\cancel{x+2}}\right)\cancel{(x+2)}(x-2)}$$
Now we have:
$$x(x-2)$$
$$=x^2-2x$$
$$\displaystyle \boxed{\therefore \dfrac{x}{x+2} \div \dfrac{1}{x^2-4}=x^2-2x}$$
