Question

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

• A. \( 2 \left[ \frac{1+\sqrt{x}}{2020} - \frac{1}{2021(1+\sqrt{x})} \right] + C \)
• B. \( 2 \left[ \frac{(1+\sqrt{x})^{-2020}}{2020} - \frac{(1+\sqrt{x})^{-2021}}{2021} \right] + C \)
• C. \(\frac{2}{2021(1 + \sqrt{x})^{2021}} - \frac{1}{1010(1 + \sqrt{x})^{2020}} + C\)
• D. \( \frac{2}{(1+\sqrt{x})^{2022}} \left[ \frac{1+\sqrt{x}}{2020} - \frac{1}{2021} \right] + C \) (This has the original term outside)

Hint:

For integrals of the form \( \int f(\sqrt{x}) dx \), the substitution \(u = \sqrt{x} \implies x=u^2 \implies dx=2udu\) is often useful.
If a further function of \(u\) arises, like \(1+u\), then a second substitution \(v=1+u\) can simplify.
Standard integral: \( \int u^n du = \frac{u^{n+1}}{n+1} + C \) (for \(n \neq -1\)).
Discrepancies with MCQ options can indicate complex forms or errors in the question/options.

Answer 1

The Correct Option is C

Integral Solution

We want to solve the integral:

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

Step 1: Substitution

Let:

\[ u = 1 + \sqrt{x} \quad \Rightarrow \quad \sqrt{x} = u - 1 \quad \Rightarrow \quad x = (u - 1)^2 \]

Differentiating both sides:

\[ dx = 2(u - 1) \, du \]

Step 2: Rewrite the Integral

Substitute into the integral:

\[ \int \frac{dx}{(1 + \sqrt{x})^{2022}} = \int \frac{2(u - 1)}{u^{2022}} \, du \]

\[ = \int \left( \frac{2u}{u^{2022}} - \frac{2}{u^{2022}} \right) du = \int \left( 2u^{-2021} - 2u^{-2022} \right) du \]

Step 3: Integrate Term by Term

\[ \int 2u^{-2021} \, du = \frac{2u^{-2020}}{-2020} = -\frac{1}{1010}u^{-2020} \]

\[ \int -2u^{-2022} \, du = \frac{2u^{-2021}}{2021} \]

Step 4: Final Answer

Substitute back \( u = 1 + \sqrt{x} \):

\[ \int \frac{dx}{(1 + \sqrt{x})^{2022}} = \frac{2}{2021(1 + \sqrt{x})^{2021}} - \frac{1}{1010(1 + \sqrt{x})^{2020}} + C \]