If int frac{dx}{x^{2022}(1+x^{2022})^{1/2022} | vedclass

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(A) Let $I = \int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}}$.Factor out $x^{2022}$ from the parenthesis:$I = \int \frac{dx}{x^{2022} \cdot (x^{2022}(x^

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(A) Let $I = \int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}}$.Factor out $x^{2022}$ from the parenthesis:$I = \int \frac{dx}{x^{2022} \cdot (x^{2022}(x^{-2022}+1))^{1/2022}} = \int \frac{dx}{x^{2022} \cdot x(1+x^{-2022})^{1/2022}} = \int \frac{dx}{x^{2023}(1+x^{-2022})^{1/2022}}$.Let $t = 1+x^{-2022}$. Then $dt = -2022x^{-2023} dx$,which implies $\frac{dx}{x^{2023}} = -\frac{1}{2022} dt$.Substituting these into the integral:$I = \int -\frac{1}{2022} t^{-1/2022} dt = -\frac{1}{2022} \cdot \frac{t^{1-1/2022}}{1-1/2022} + C = -\frac{1}{2022} \cdot \frac{t^{2021/2022}}{2021/2022} + C = -\frac{t^{2021/2022}}{2021} + C$.Substituting $t = 1+x^{-2022} = \frac{x^{2022}+1}{x^{2022}}$:$I = -\frac{(\frac{x^{2022}+1}{x^{2022}})^{2021/2022}}{2021} + C = -\frac{(1+x^{2022})^{2021/2022}}{2021 \cdot (x^{2022})^{2021/2022}} + C = -\frac{(1+x^{2022})^{2021/2022}}{2021 x^{2021}} + C$.Comparing this with $\frac{-(1+x^m)^{n/m}}{nx^n} + C$,we get $m=2022$ and $n=2021$.Thus,$m-n = 2022-2021 = 1$.

If $\int \frac{dx}{x^{2022}(1+x^{2022})^{1/2022}} = \frac{-(1+x^m)^{n/m}}{nx^n} + C$,then $m-n=$

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