Question:
\[ \int \frac{\mathrm{d} x}{\sqrt{x}+x}= \]
\[ \int \frac{\mathrm{d} x}{\sqrt{x}+x}= \]
Show Hint
Substitute t = sqrt(x) so that the factor t cancels.
Updated On: Nov 13, 2026
- \(\mathrm{log}\sqrt{x}+c\), where \(c\) is the constant of integration.
- \(\mathrm{log}(\sqrt{x}+x)+\mathrm{c}\), where c is the constant of integration.
- \(\mathrm{log}(1+\sqrt{x})+\mathrm{c}\), where c is the constant of integration.
- \(2\mathrm{log}(1+\sqrt{x})+\mathrm{c}\), where c is the constant of integration.
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Put \(\sqrt{x} = t\), so \(x = t^2\) and \(dx = 2t\,dt\).
Step 2: Substitute
\(\displaystyle\int\dfrac{dx}{\sqrt{x} + x} = \int\dfrac{2t\,dt}{t + t^2} = \int\dfrac{2\,dt}{1 + t}\).
Step 3: Integrate
\(= 2\log|1 + t| + c = 2\log(1 + \sqrt{x}) + c\).
Final Answer:
The result is \(2\log(1+\sqrt{x}) + c\). \[ \boxed{\text{(D) }2\log(1+\sqrt{x}) + c} \]
Put \(\sqrt{x} = t\), so \(x = t^2\) and \(dx = 2t\,dt\).
Step 2: Substitute
\(\displaystyle\int\dfrac{dx}{\sqrt{x} + x} = \int\dfrac{2t\,dt}{t + t^2} = \int\dfrac{2\,dt}{1 + t}\).
Step 3: Integrate
\(= 2\log|1 + t| + c = 2\log(1 + \sqrt{x}) + c\).
Final Answer:
The result is \(2\log(1+\sqrt{x}) + c\). \[ \boxed{\text{(D) }2\log(1+\sqrt{x}) + c} \]
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