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The value of ∫(x+1x)2dx\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 dx is:

Solution

✅ Correct Option: 2

Expand the square: (x+1/x)2=x+2+1/x(\sqrt{x} + 1/\sqrt{x})^2 = x + 2 + 1/x. Integrate term by term: ∫(x+2+1/x)dx=x2/2+2x+log⁡∣x∣+C\int(x + 2 + 1/x)dx = x^2/2 + 2x + \log|x| + C.

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