int frac{d x}{(1+sqrt{x})^{2022}} = | vedclass

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Question

(A) Let $I = \int \frac{d x}{(1+\sqrt{x})^{2022}}$.Substitute $x = t^2$,then $dx = 2t dt$.The integral becomes $I = \int \frac{2t dt}{(1+t)^{2022}}$.R

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{d x}{(1+\sqrt{x})^{2022}}$.Substitute $x = t^2$,then $dx = 2t dt$.The integral becomes $I = \int \frac{2t dt}{(1+t)^{2022}}$.Rewrite the numerator as $t = (t+1) - 1$:$I = 2 \int \frac{t+1-1}{(1+t)^{2022}} dt = 2 \left[ \int \frac{t+1}{(1+t)^{2022}} dt - \int \frac{1}{(1+t)^{2022}} dt \right]$.$I = 2 \left[ \int \frac{1}{(1+t)^{2021}} dt - \int \frac{1}{(1+t)^{2022}} dt \right]$.Integrating both terms:$I = 2 \left[ \frac{(1+t)^{-2020}}{-2020} - \frac{(1+t)^{-2021}}{-2021} \right] + C$.$I = 2 \left[ \frac{-1}{2020(1+t)^{2020}} + \frac{1}{2021(1+t)^{2021}} \right] + C$.Factor out $\frac{1}{(1+t)^{2021}}$:$I = \frac{2}{(1+t)^{2021}} \left[ \frac{-(1+t)}{2020} + \frac{1}{2021} \right] + C$.Substituting $t = \sqrt{x}$ back:$I = \frac{2}{(1+\sqrt{x})^{2021}} \left[ \frac{-(1+\sqrt{x})}{2020} + \frac{1}{2021} \right] + C$.

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

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