For $x>e,\int\frac{dx}{x-\sqrt{x}}$ 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 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 $t=\sqrt{x}$:
$2\ln(\sqrt{x}-1) + C$