\int {d x}/{(1+\sqrt{x})^{2022}}=

Correct Answer is : 2(1+√x)2021[1+√x2020−12021]+C

{2}/{(1+\sqrt{x})^{2021}}[\frac{1+\sqrt{x}{2020}-\frac{1}{2021}]+C

{2}/{(1+\sqrt{x})^{2022}}[\frac{1+\sqrt{x}{2020}-\frac\sqrt{x}{2021}]+C

\frac{2}{(1+\sqrt{x})}[ {(1+\sqrt{x})^{2022}}/{2022}- {(1+\sqrt{x})^{2021}}/{2021}]+C

{1}/{(1+\sqrt{x})^{2}}[ {1}/{(1+\sqrt{x})^{1010}}- {1}/{(1+\sqrt{x})^{1011}}]+C

AP EAPCET 07-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 70 of 160

Marks: +1, -0

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(1+√x)2022

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