Question:
Compute \( \int \frac{\sqrt{x + 1}}{\sqrt{x}} \, dx \)
Compute \( \int \frac{\sqrt{x + 1}}{\sqrt{x}} \, dx \)
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When handling integrals with square roots, split the terms to simplify the integrand and integrate each term separately.
Updated On: Apr 18, 2026
- \( 2\sqrt{x+1} - \sqrt{x} + C \)
- \( 2\sqrt{x+1} + \sqrt{x} + C \)
- \( \sqrt{x+1} - \sqrt{x} + C \)
- \( \sqrt{x+1} + \sqrt{x} + C \)
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The Correct Option is A
Solution and Explanation
Step 1: Simplify the integrand.
We can rewrite the integrand as: \[ \int \frac{\sqrt{x+1}}{\sqrt{x}} \, dx = \int \frac{\sqrt{x} + \sqrt{1}}{\sqrt{x}} \, dx = \int \left( 1 + \frac{1}{\sqrt{x}} \right) \, dx \]
Step 2: Integrate each term.
The integral of \( 1 \) is: \[ \int 1 \, dx = x \] The integral of \( \frac{1}{\sqrt{x}} \) is: \[ \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} \]
Step 3: Combine the results.
Thus, the total integral is: \[ x + 2\sqrt{x} + C \]
Final Answer: \( 2\sqrt{x+1} - \sqrt{x} + C \).
We can rewrite the integrand as: \[ \int \frac{\sqrt{x+1}}{\sqrt{x}} \, dx = \int \frac{\sqrt{x} + \sqrt{1}}{\sqrt{x}} \, dx = \int \left( 1 + \frac{1}{\sqrt{x}} \right) \, dx \]
Step 2: Integrate each term.
The integral of \( 1 \) is: \[ \int 1 \, dx = x \] The integral of \( \frac{1}{\sqrt{x}} \) is: \[ \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} \]
Step 3: Combine the results.
Thus, the total integral is: \[ x + 2\sqrt{x} + C \]
Final Answer: \( 2\sqrt{x+1} - \sqrt{x} + C \).
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