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

Looking at the denominator $x - \sqrt{x}$, factor out $\sqrt{x}$:

$x - \sqrt{x}$

$= \sqrt{x} \cdot \sqrt{x} - \sqrt{x}$

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

The integral becomes:

$\int \frac{dx}{\sqrt{x}(\sqrt{x} - 1)}$

---

Let $u = \sqrt{x}$

Then $x = u^2$ and $dx = 2u \, du$

Substituting:

$\int \frac{2u \, du}{u(u - 1)}$

---

Canceling $u$ from numerator and denominator:

$\int \frac{2u \, du}{u(u - 1)}$

$= \int \frac{2 \, du}{u - 1}$

---

Using the standard integral formula $\int \frac{1}{u-1} du = \ln|u-1| + C$:

$\int \frac{2 \, du}{u - 1}$

$= 2 \ln|u - 1| + C$

---

Substituting back $u = \sqrt{x}$:

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

Therefore, $\int \frac{dx}{x - \sqrt{x}} = 2 \log_e |\sqrt{x} - 1| + C$