If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals: 
$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$
First simplify $f(x+2)$:
\[f(x+2)=\frac{(x+2)(x+1)}{2}.\]
Then simplify the answers and see which one matches.
$f(x)+f(2)=\frac{x(x-1)}{2} + 1,$
$(x+2)f(x)=\frac{(x+2)x(x-1)}{2},$
$x(x+2)f(x)=\frac{(x+2)x^2(x-1)}{2},$
$\frac{xf(x)}{x+2}=\frac{x^2(x-1)}{2(x+2)},$
$\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}.$
Since $f(x+2)=\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}$, the answer is $\boxed{\textbf{(E)}\ \frac{(x+2)f(x+1)}{x}}.$