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

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$\frac{2 {(1 + \sqrt{x})}^{- 2019}}{2019} + \frac{{(1 + \sqrt{x})}^{- 2018}}{1009} + C$

$\frac{1}{2019 {(1 + \sqrt{x})}^{2019}} + \frac{1}{1009 (1 + \sqrt{x})} + c$

$\frac{2 {(1 + \sqrt{x})}^{- 2019}}{2019} - \frac{{(1 + \sqrt{x})}^{- 2018}}{1009} + C$

$\frac{{(1 + \sqrt{x})}^{- 2019}}{2019} - \frac{{(1 + \sqrt{x})}^{- 2018}}{1009} + c$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \frac{1}{(1 + \sqrt{x})^{2020}} d x = \int \frac{\sqrt{x}}{\sqrt{x} (1 + \sqrt{x})^{2020}} d x \\ Let 1 + \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \\ = 2 \int \frac{1}{t^{2019}} d t - 2 \int \frac{1}{t^{2020}} d t \\ = \frac{- 2}{2018 t^{2018}} + \frac{2}{2019 t^{2019}} + C \end{array}$