Answer
$=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}, \quad x\neq-2,2$
Work Step by Step
The denominators are different.
Find a common denominator.
LCD=$(x+2)(x-2)$
$\displaystyle \frac{2x}{x+2}+\frac{x+2}{x-2}= \displaystyle \frac{2x(x-2)}{(x+2)(x-2)}+\frac{(x+2)(x+2)}{(x-2)(x+2)}$
$=\displaystyle \frac{2x^{2}-4x+x^{2}+4x+4}{(x+2)(x-2)}$
$=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}$
... exclude values that yield 0 in the denominator:
$x\neq-2,2$
... and, cancel common factors (here, there are none)
$=\displaystyle \frac{3x^{2}+4}{(x+2)(x-2)}, \quad x\neq-2,2$
