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$\int \frac{d x}{(1+\sqrt{x})^{2022}}=$
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The correct answer is:
$\frac{2}{(1+\sqrt{x})^{2021}}\left[\frac{-(1+\sqrt{x})}{2020}+\frac{1}{2021}\right]+C$
$\begin{aligned} I & =\int \frac{d x}{(1+\sqrt{x})^{2022}}, \text { put } x=t^2 \Rightarrow d x=2 t d t \\ \Rightarrow & \int \frac{2 t d t}{(1+t)^{2022}} \\ & =2\left[\int \frac{t+1}{(1+t)^{2022}} d t-\int \frac{1}{(1+t)^{2022}} d t\right] \\ & =2\left[\int \frac{1}{(t+1)^{2021}} d t-\int \frac{1}{(1+t)^{2022}} d t\right] \\ & =2\left[\frac{-1}{\left.(2020) \cdot(t+1)^{2020}+\frac{1}{(2021) \cdot(t+1)^{2021}}\right]+C}\right. \\ \therefore I= & \frac{2}{(1+\sqrt{x})^{2021}}\left[\frac{-(1+\sqrt{x})}{2020}+\frac{1}{2021}\right]+C\end{aligned}$
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