Find the following integral: $\int(1-x) \sqrt{x} \, dx$

To evaluate the integral $\int(1-x) \sqrt{x} \, dx$,we first distribute $\sqrt{x}$ into the parentheses:
$= \int (\sqrt{x} - x \cdot \sqrt{x}) \, dx$
$= \int (x^{\frac{1}{2}} - x^1 \cdot x^{\frac{1}{2}}) \, dx$
$= \int (x^{\frac{1}{2}} - x^{\frac{3}{2}}) \, dx$
Now,apply the power rule for integration,$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:
$= \int x^{\frac{1}{2}} \, dx - \int x^{\frac{3}{2}} \, dx$
$= \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} - \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + C$
$= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C$
$= \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + C$
where $C$ is an arbitrary constant.