\( \int\left(\sqrt{x}-\displaystyle\frac{1}{\sqrt{x}}\right)^{2}dx \) = (where \( x\gt 0 \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{x^{2}}{2}+\log_{e}|x|-2x+c \) : where \( c \) is an arbitrary constant
The square opens into three simple terms:
\( \displaystyle \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}=x-2+\frac{1}{x} \)
Integrating term by term,
\( \displaystyle \int\left(x-2+\frac{1}{x}\right)dx=\frac{x^{2}}{2}-2x+\log_{e}|x|+c \)
NCERT (2026-27 edition):
Integration as an Inverse Process of Differentiation
