Correct option is 421+x−20192019−1+x−20181009+C

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$\int \frac{1}{{(1 + \sqrt{x})}^{2020}} d x =$

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120191+x2019+110091+x+c

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detailed solution

Correct option is D

I=∫1(1+x)2020dx=∫xx(1+x)2020dx Let 1+x=t⇒12xdx=dt=2∫1t2019dt−2∫1t2020dt=−22018t2018+22019t2019+C

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∫11+x2020dx=

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$\int \frac{1}{{(1 + \sqrt{x})}^{2020}} d x =$