CUET · MATHS · PYQ PAPER 2023
The value of \(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 d x\) is:
- A \(\frac{x^2}{2}+\log |x|+C\), (where C is constant of integration)
- B \(\frac{x^2}{2}+\log |x|+2 x+C\), (where \(C\) is constant of integration)
- C \(\frac{3}{2}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+C\), (where C is constant of integration)
- D \(\frac{2}{3} x^{\frac{3}{2}}+2 \sqrt{x}+C\), (where C is constant of integration)
Answer & Solution
Correct Answer
(B) \(\frac{x^2}{2}+\log |x|+2 x+C\), (where \(C\) is constant of integration)
Step-by-step Solution
Detailed explanation
\(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 d x = \int\left(x+2+\frac{1}{x}\right) d x\) \(= \frac{x^2}{2}+2 x+\log |x|+C\)
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