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AP EAPCET 06-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

\int\frac{dx}{x^{2022}(1+x^{2022})^{\frac{1}{2022}}}=-\frac{(1+x^{m})^{\frac{n}{m}}}{n x^{n}}+C

m-n=

Solution: 👈: Video Solution

Given

\int\frac{dx}{x^{2022}(1+x^{2022})^{\frac{1}{2022}}}=-\frac{(1+x^{m})^{\frac{n}{m}}}{n x^{n}}+C

Let

I=\int\frac{dx}{x^{2022}(1+x^{2022})^{\frac{1}{2022}}}

\;=\int\frac{dx}{x^{2022}\left[x^{2022}\left(1+\frac{1}{x^{2022}}\right)\right]^{\frac{1}{2022}}}

\;=\int\frac{dx}{x^{2023}\left[1+\frac{1}{x^{2022}}\right]^{\frac{1}{2022}}}

Let

\;1+\frac{1}{x^{2022}}=t

\Rightarrow -2022 \times \frac{1}{x^{2023}}dx=dt

\Rightarrow \;\; \frac{1}{x^{2023}}dx=-\frac{1}{2022}dt

\Rightarrow \;\; I=\frac{-1}{2022}\int\frac{1}{t^{\frac{1}{2022}}}dt

\;=\frac{-1}{2022}\int t^{-\frac{1}{2022}}dt

\;=-\frac{1}{2022}\left(\frac{t^{-\frac{1}{2022}+1}}{1}\right)+C

=-\frac{1}{2021}\left[(1+x^{-2022})^{\frac{2021}{2022}}\right]

\;=-\frac{1}{2021}\left[\frac{x^{2022}+1}{x^{2022}}\right]^{\frac{2021}{2022}}

\;=\frac{-[x^{2022}+1]^{2021}}{x^{2021}}\times 2021

\;\therefore n=2021 \text{ and } m=2022

\;\therefore m-n=1

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