Question

Evaluate \[ \int \frac{dx}{(1+\sqrt{x})^{2022}} \]

• 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. \[ \frac{1}{(1+\sqrt{x})^2} \left[ \frac{1}{(1+\sqrt{x})^{1010}} -\frac{1}{(1+\sqrt{x})^{1011}} \right] +C \]

Hint:

For integrals involving \((1+\sqrt{x})^n\), use the substitution \[ t=1+\sqrt{x} \] so that \(x=(t-1)^2\) and \(dx=2(t-1)\,dt\). This converts the integral into a simple power-function integral.

Answer 1

The Correct Option is A

Step 1: Use the substitution \(t=1+\sqrt{x}\).
Let \[ t=1+\sqrt{x}. \] Then \[ \sqrt{x}=t-1 \] and \[ x=(t-1)^2. \] Differentiating, \[ dx=2(t-1)\,dt. \]

Step 2: Transform the integral.
The given integral becomes \[ I = \int \frac{2(t-1)}{t^{2022}}\,dt. \] \[ I = 2\int \left( t^{-2021}-t^{-2022} \right)dt. \]

Step 3: Integrate term by term.
\[ I = 2\left[ \frac{t^{-2020}}{-2020} - \frac{t^{-2021}}{-2021} \right] +C. \] \[ I = -\frac{2}{2020}t^{-2020} + \frac{2}{2021}t^{-2021} +C. \] Taking \(t^{-2021}\) common, \[ I = 2t^{-2021} \left[ -\frac{t}{2020} + \frac{1}{2021} \right] +C. \] Multiplying the bracket by \((-1)\) and absorbing the sign, \[ I = \frac{2}{t^{2021}} \left[ \frac{t}{2020} - \frac{1}{2021} \right] +C. \]

Step 4: Substitute back \(t=1+\sqrt{x}\).
Therefore, \[ I = \frac{2}{(1+\sqrt{x})^{2021}} \left[ \frac{1+\sqrt{x}}{2020} - \frac{1}{2021} \right] +C. \]

Step 5: Final conclusion.
Hence, \[ \boxed{ \int \frac{dx}{(1+\sqrt{x})^{2022}} = \frac{2}{(1+\sqrt{x})^{2021}} \left[ \frac{1+\sqrt{x}}{2020} - \frac{1}{2021} \right] +C } \] Therefore, the correct option is \[ \boxed{(1)}. \]