For example, the remainder when $16 x^{5}+x+1$ is divided by $x^{2}+1$ is $\qquad$
To find the remainder, we can first express $x^{5}+x+1$ using the factor $x^{2}+1$. By the properties of polynomials, we can assume
$$
\begin{array}{l}
x^{5}+x+1 \\
=A x\left(x^{2}+1\right)^{2}+B x\left(x^{2}+1\right)+C x+D \\
=A x^{5}+(2 A+B) x^{3}+(A+B+C) x \\
\quad+D .
\end{array}
$$

By comparing the coefficients of corresponding terms on both sides of the equation, we get
$$
\left\{\begin{array}{l}
A=1, \\
2 A+B=0, \\
A+B+C=1, \\
D=1 .
\end{array}\right.
$$

Solving these equations, we find $A=1, B=-2, C=2, D=1$.
$$
\begin{aligned}
\therefore & x^{5}+x+1 \\
& =x\left(x^{2}+1\right)^{2}-2 x\left(x^{2}+1\right)+2 x+1 .
\end{aligned}
$$

Since terms containing $x^{2}+1$ can all be divided by $x^{2}+1$, the remainder is $2 x+1$.