If $\int \frac{d x}{x^{2022}\left(1+x^{2022}\right)^{1 / 2022}}=\frac{-\left(1+x^m\right)^{n / m}}{n x^n}+C$…
If $\int \frac{d x}{x^{2022}\left(1+x^{2022}\right)^{1 / 2022}}=\frac{-\left(1+x^m\right)^{n / m}}{n x^n}+C$, then $\mathrm{m}-\mathrm{n}=$
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Solution
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
$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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