Given $\int \frac{\mathrm{dx}}{\mathrm{x}^{2022}\left(1+\mathrm{x}^{2022}\right)^{1 / 2002}}$
$$
\Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}^{2023}\left(1+\mathrm{x}^{-2022}\right)^{1 / 2022}}
$$
Let $1+\mathrm{x}^{-2022}=\mathrm{t}$
$$
\Rightarrow-2022 \mathrm{x}^{-2003} \mathrm{dx}=\mathrm{dt}
$$
or $\frac{\mathrm{dx}}{\mathrm{x}^{2023}}=\frac{-1}{2022} \mathrm{dt}$
$$
\begin{aligned}
& =\int \frac{-\mathrm{dt}}{2002}(\mathrm{t}) \frac{1}{2022} \\
& =\frac{-1}{2022} \int \mathrm{t}^{\frac{-1}{2002}} \mathrm{dt}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \frac{-1}{2022} \frac{\mathrm{t}^{\frac{-1}{2022}+1}}{\frac{-1}{2022}+1} \\
& \Rightarrow \frac{-1}{2022} \frac{\left(1+\mathrm{x}^{2002}\right)^{\frac{2001}{2022}}}{\frac{2021}{2022}}
\end{aligned}
$$
$$
\Rightarrow \frac{-1}{2022} \frac{\left(1+\mathrm{x}^{2002}\right)^{\frac{2001}{2022}}}{\frac{2021}{2022}}
$$
$$
\frac{\left(1+x^{2022}\right)^{\frac{2021}{2002}}}{{ }^{2021} x^{2021}}+c
$$
$\Rightarrow \mathrm{m}=2022$ and $\mathrm{n}=2021$
$$
\Rightarrow \mathrm{m}-\mathrm{n}=1
$$