For $x>e,\int\frac{dx}{x-\sqrt{x}}$ is equal to

$2\log_e|\sqrt{x}-1|+C$: C is an arbitray constant

The correct answer is Option (4) → $2\log_e|\sqrt{x}-1|+C$: C is an arbitray constant

Evaluate:

$\displaystyle \int \frac{dx}{x - \sqrt{x}}$

Factor denominator:

$x - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$

Rewrite the integrand:

$\frac{1}{x - \sqrt{x}} = \frac{1}{\sqrt{x}(\sqrt{x} - 1)} = \frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x}-1}$

Let $t = \sqrt{x}$, so $x = t^{2}$ and $dx = 2t\,dt$.

Substitute:

$\int \frac{dx}{x-\sqrt{x}} = \int \frac{2t}{t(t-1)}\,dt = \int \frac{2}{t-1}\,dt$

Integrate:

$2\ln|t-1| + C$

Put back $t=\sqrt{x}$:

$2\ln(\sqrt{x}-1) + C$