Q110 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

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

  • Correct answer
    A
    $\frac{2}{(1+\sqrt{x})^{2021}}\left[\frac{-(1+\sqrt{x})}{2020}+\frac{1}{2021}\right]+C$
  • B
    $\frac{2}{(1+\sqrt{x})^{2022}}\left[\frac{1+\sqrt{x}}{2020}-\frac{\sqrt{x}}{2021}\right]+C$
  • C
    $\frac{2}{(1+\sqrt{x})}\left[\frac{(1+\sqrt{x})^{2022}}{2022}-\frac{(1+\sqrt{x})^{2021}}{2021}\right]+C$
  • D
    $\frac{1}{(1+\sqrt{x})^2}\left[\frac{1}{(1+\sqrt{x})^{1010}}-\frac{1}{(1+\sqrt{x})^{1011}}\right]+C$

Solution

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

Hence, the correct answer is option: 'A'.

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