$\int \dfrac{1}{x \sqrt{a x-x^2}} d x$ is

Let $I = \int \dfrac{1}{x \sqrt{ax - x^2}} dx$.

Rewrite the integral as $I = \int \dfrac{1}{x \sqrt{x(a - x)}} dx$.

Divide the numerator and denominator inside the square root by $x^2$: $I = \int \dfrac{1}{x \cdot x \sqrt{\dfrac{a - x}{x}}} dx = \int \dfrac{1}{x^2 \sqrt{\dfrac{a}{x} - 1}} dx$.

Let $t = \sqrt{\dfrac{a}{x} - 1}$. Then $t^2 = \dfrac{a}{x} - 1$, which implies $t^2 + 1 = \dfrac{a}{x}$, so $x = \dfrac{a}{t^2 + 1}$.

Differentiating $x$ with respect to $t$: $dx = a \cdot (-1) (t^2 + 1)^{-2} (2t) dt = \dfrac{-2at}{(t^2 + 1)^2} dt$.

Substitute $x$ and $dx$ into the integral:

$I = \int \dfrac{1}{\left( \dfrac{a}{t^2 + 1} \right)^2 \cdot t} \cdot \left( \dfrac{-2at}{(t^2 + 1)^2} \right) dt$.

$I = \int \dfrac{(t^2 + 1)^2}{a^2 \cdot t} \cdot \dfrac{-2at}{(t^2 + 1)^2} dt = \int \dfrac{-2}{a} dt$.

$I = -\dfrac{2}{a} t + C$.

Substituting back $t = \sqrt{\dfrac{a - x}{x}}$:

$I = -\dfrac{2}{a} \sqrt{\dfrac{a - x}{x}} + C$.

Answer: $\dfrac{-2}{a} \sqrt{\dfrac{a-x}{x}}+C$