If $f(x)=\frac{x^4+x^2}{x+1}$, then $f(i)$, where $i=\sqrt{-1}$, is equal to
$\text{(A) } 1+i\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 0\quad \text{(E) } -1-i$
$i^4 = 1$ and $i^2=-1$, so the numerator is $0$. As long as the denominator is not $0$, which it isn't, the answer is $0 \Rightarrow$ $\fbox{D}$