Question

$\int \frac{dx}{\sqrt{x+1} + \sqrt{x}} =$

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

Hint:

Rationalization isn't just for limits; it's a powerful tool for simplifying radical integrals too.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To integrate a fraction with a sum of square roots in the denominator, the most effective method is to rationalize the denominator. This converts the expression into a simpler form where basic integration rules can be applied.
Step 2: Key Formula or Approach:
1. Rationalization: Multiply the numerator and denominator by the conjugate \(\sqrt{x+1} - \sqrt{x}\).
2. Power rule for integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + c\).
Step 3: Detailed Explanation:
First, rationalize the denominator: \[ \frac{1}{\sqrt{x+1} + \sqrt{x}} \times \frac{\sqrt{x+1} - \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} = \frac{\sqrt{x+1} - \sqrt{x}}{(x+1) - x} = \sqrt{x+1} - \sqrt{x} \] The integral becomes: \[ \int (\sqrt{x+1} - \sqrt{x}) \, dx = \int (x+1)^{1/2} \, dx - \int x^{1/2} \, dx \] Apply the power rule to each term: \[ = \frac{(x+1)^{3/2}}{3/2} - \frac{x^{3/2}}{3/2} + c \] \[ = \frac{2}{3}(x+1)^{3/2} - \frac{2}{3}x^{3/2} + c = \frac{2}{3} \left[ (x+1)^{3/2} - x^{3/2} \right] + c \]
Step 4: Final Answer:
The integral is \( \frac{2}{3} \left[ (x+1)^{3/2} - x^{3/2} \right] + c \).