Let $I=\int \frac{1}{x \sqrt{a x-x^2}} d x$
Putting $x=\frac{a}{t}$
$$
\begin{aligned}
& \Rightarrow \quad d x=\frac{-a}{t^2} d t \\
& \therefore \quad I=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-\frac{a^2}{t^2}}} \times-\frac{a}{t^2} d t
\end{aligned}
$$

$$
\begin{aligned}
& =\int \frac{-t}{t^2 \sqrt{\frac{a^2}{t}-\frac{a^2}{t^2}}} d t \\
& =\int \frac{-1}{a t \sqrt{\frac{1}{t}-\frac{1}{t^2}}} d t \\
& =\frac{-1}{a} \int \frac{1}{\sqrt{\frac{t^2}{t}-\frac{t^2}{t^2}}} d t \\
& =\frac{-1}{a} \int \frac{1}{\sqrt{t-1}} d t=\frac{-1}{a} \int(t-1)^{-1 / 2} d t \\
& =\frac{-1}{a} \frac{(t-1)^{1 / 2}}{1 / 2}+C=\frac{-2}{a}\left(\frac{a}{x}-1\right)^{1 / 2}+C \\
& I=\frac{-2}{a} \sqrt{\frac{a-x}{x}+C} .
\end{aligned}
$$
Option (3) is correct.