Question

\[ \int \frac{dx}{\sqrt{x+1}+\sqrt{x}}= \]

• A. \(\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c\)
• B. \(\frac{2}{3}\left[(x+1)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]+c\)
• C. \(\frac{3}{2}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c\)
• D. \(\frac{3}{2}\left[(x+1)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]+c\)

Hint:

For integrals with \(\sqrt{x+1}+\sqrt{x}\) in the denominator, rationalize using \(\sqrt{x+1}-\sqrt{x}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are asked to integrate a function with a sum of square roots in the denominator.
This is a classic problem that is solved by rationalizing the denominator, which converts the complex denominator into a simple expression (usually a constant).
Step 2: Key Formula or Approach:
1. Rationalization: multiply numerator and denominator by \( \sqrt{x+1} - \sqrt{x} \).
2. Power rule for integration: \( \int x^n dx = \frac{x^{n+1}}{n+1} + c \).
Step 3: Detailed Explanation:

Rationalizing the denominator:
\[ \frac{1}{\sqrt{x+1} + \sqrt{x}} \cdot \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} = \frac{\sqrt{x+1} - \sqrt{x}}{(x+1) - (x)} \]
\[ = \frac{\sqrt{x+1} - \sqrt{x}}{1} = \sqrt{x+1} - \sqrt{x} \]

Setting up the integral:
\[ \int (\sqrt{x+1} - \sqrt{x}) dx = \int (x+1)^{1/2} dx - \int x^{1/2} dx \]

Integrating each term:
For the first term, let \( u = x+1, du = dx \):
\[ \int (x+1)^{1/2} dx = \frac{(x+1)^{3/2}}{3/2} = \frac{2}{3}(x+1)^{3/2} \]
For the second term:
\[ \int x^{1/2} dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \]

Combining the results:
\[ \frac{2}{3}(x+1)^{3/2} - \frac{2}{3}x^{3/2} + c = \frac{2}{3} [(x+1)^{3/2} - (x)^{3/2}] + c \]

Step 4: Final Answer:
The integral is \( \frac{2}{3} [(x+1)^{3/2} - (x)^{3/2}] + c \).