Answer
$\frac{3x^{2}+4}{(x+2)(x-2)}; x \ne 2, -2;$
Work Step by Step
$\frac{2x}{(x+2)} + \frac{(x+2)}{(x-2)}$
$= \frac{2x(x-2)+(x+2)(x+2)}{(x+2)(x-2)}; x \ne 2, -2;$
$= \frac{2x(x-2)+(x+2)^{2}}{(x+2)(x-2)}; x \ne 2, -2;$
$[(A+B)^{2} = A^{2}+2AB+B^{2}$ So,
$(x+2)^{2} = x^{2}+4x+4] $
$= \frac{2x^{2}-4x+x^{2}+4x+4}{(x+2)(x-2)} ; x \ne 2, -2;$
Combine like terms.
$\frac{3x^{2}+4}{(x+2)(x-2)}; x \ne 2, -2;$
