Using long division,find the quotient and the remainder when the polynomial $x^{4}+1$ is divided by $x+1$.

(N/A) To divide $x^{4}+1$ by $x+1$,we perform long division:
$x+1$ is the divisor and $x^{4}+0x^{3}+0x^{2}+0x+1$ is the dividend.
$1$. Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(x)$ to get $x^{3}$.
$2$. Multiply $x^{3}(x+1) = x^{4}+x^{3}$ and subtract from the dividend: $(x^{4}+0x^{3}) - (x^{4}+x^{3}) = -x^{3}$.
$3$. Bring down the next term $(0x^{2})$ to get $-x^{3}+0x^{2}$.
$4$. Divide $-x^{3}$ by $x$ to get $-x^{2}$. Multiply $-x^{2}(x+1) = -x^{3}-x^{2}$ and subtract: $(-x^{3}+0x^{2}) - (-x^{3}-x^{2}) = x^{2}$.
$5$. Bring down $0x$ to get $x^{2}+0x$. Divide $x^{2}$ by $x$ to get $x$. Multiply $x(x+1) = x^{2}+x$ and subtract: $(x^{2}+0x) - (x^{2}+x) = -x$.
$6$. Bring down $1$ to get $-x+1$. Divide $-x$ by $x$ to get $-1$. Multiply $-1(x+1) = -x-1$ and subtract: $(-x+1) - (-x-1) = 2$.
Thus,the quotient is $x^{3}-x^{2}+x-1$ and the remainder is $2$.