The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$  equals:
$\textbf{(A)}\ 2\qquad  \textbf{(B)}\ 2 - \sqrt{2}\qquad  \textbf{(C)}\ 2 + \sqrt{2}\qquad  \textbf{(D)}\ 2\sqrt{2}\qquad  \textbf{(E)}\ \frac{\sqrt{2}}{2}$
To make this problem easier to solve, lets get the radicals out of the denominator. For $\frac{1}{2 + \sqrt2}$, we will multiply the numerator and denominator by $2- \sqrt2$ so,
$\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}$.
Now, the other fraction we need to get the radical out of the denominator is $\frac{1}{\sqrt2 - 2}$. Here, we will  multiply by the conjugate again, $\sqrt2 + 2$. So that simplifies to
$\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}$.
So now our simplified equation is
$2 + \sqrt2 + \frac{2- \sqrt2}{2} + \frac{\sqrt2 + 2}{-2} \Rightarrow 2+ \sqrt2 + \frac{2 - \sqrt2}{2} - \frac{\sqrt2 +2}{2}$
Bringing everything to the same denominator and combining like terms, we get
$\frac{4 + 2\sqrt2 + 2 - \sqrt2 - \sqrt2 - 2}{2} \Rightarrow \frac{4}{2} \Rightarrow 2 \Rightarrow \boxed{A}$