The given expression can be simplified by finding the common denominator and adding the terms inside the parentheses.

First, let's find the common denominator for the terms inside the parentheses:

The denominator of the first term is x^2 - 4, which can be factored as $x+2$$x-2$.The denominator of the second term is 2x - x^2, which can be rewritten as -x^2 + 2x.

Now, the common denominator for both terms is $x+2$$x-2$$-x^2+2x$.

Next, rewrite each term with the common denominator and then add them together:

$\frac{2}{x^2-4}$ can be rewritten as $\frac{2}{(x+2)(x-2)}$.$\frac{1}{2x-x^2}$ can be rewritten as $\frac{1}{(-x^2+2x)}$.

So, the expression inside the parentheses becomes:
$$\frac{2}{(x+2)(x-2)}+\frac{1}{(-x^2+2x)}$$

Now, we multiply the first term by $\frac{(x+2)}{(x+2)}$ to get a common denominator:
$$\frac{2(x+2)}{(x+2)(x-2)}+\frac{1}{(-x^2+2x)}$$

Now we have:
$$\frac{2(x+2)}{(x+2)(x-2)}+\frac{1}{(-x^2+2x)}:\frac{1}{x^2+4x+4}$$

To divide by a fraction, we multiply by its reciprocal:
$$\frac{2(x+2)}{(x+2)(x-2)}+\frac{1}{(-x^2+2x)}\cdot \frac{x^2+4x+4}{1}$$

Simplifying further, we get:
$$\frac{2(x+2)}{(x+2)(x-2)}+\frac{(x^2+4x+4)}{(-x^2+2x)}$$ $$\frac{2(x+2)}{(x+2)(x-2)}+\frac{(x+2)(x+2)}{(-x)(x-2)}$$

At this step, the numerator and denominator in the second term can cancel out, leading to:
$$2(x+2)+(x+2)$$ $$2x+4+x+2$$ $$3x+6$$

Therefore, the simplified expression is $3x+6$, assuming x is not equal to 2 or 0.