For \(x \gt e,\ \int\displaystyle\frac{dx}{x-\sqrt{x}}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\log_e|\sqrt{x}-1|+C\): C is an arbitrary constant
Let \(t = \sqrt{x}\) (substitution), so \(x = t^2\) and \(dx = 2t\,dt\):
\( \displaystyle \int \frac{dx}{x - \sqrt{x}} = \int \frac{2t\,dt}{t^2 - t} \)
\( \displaystyle = \int \frac{2t\,dt}{t(t - 1)} = \int \frac{2\,dt}{t - 1} \)
\(= 2\log_e|t - 1| + C\)
\(= 2\log_e|\sqrt{x} - 1| + C\)
Shortcut: Differentiate Option (4):
\( \displaystyle \frac{d}{dx}\, 2\log_e|\sqrt{x} - 1| = \frac{2}{\sqrt{x} - 1} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x} - 1)} = \frac{1}{x - \sqrt{x}} \)
NCERT (2026-27 edition):
Integration by Substitution
