$\cfrac{1}{{x\sqrt {ax - {x^2}} }}$ [Hint : Put $x = \cfrac{a}{t}$ ]

Let $I = \int {\cfrac{1}{{x\sqrt {ax - {x^2}} }}dx}$ Put $x = \cfrac{a}{t}$ $\Rightarrow$ $dx = - \cfrac{a}{{{t^2}}}dt$

Now, $x\sqrt {ax - {x^2}} = \cfrac{a}{t}\sqrt {\cfrac{{{a^2}}}{t} - \cfrac{{{a^2}}}{{{t^2}}}} = \cfrac{{{a^2}}}{t}\sqrt {\cfrac{1}{t} - \cfrac{1}{{{t^2}}}} = \cfrac{{{a^2}}}{{{t^2}}}\sqrt {t - 1}$

$\therefore$ $I = \int {\cfrac{1}{{\cfrac{{{a^2}}}{{{t^2}}}\sqrt {t - 1} }}} \times \left( { - \cfrac{a}{{{t^2}}}} \right)dt = - \cfrac{1}{a}\int {\cfrac{1}{{\sqrt {t - 1} }}dt}$

$= - \cfrac{1}{a} \cdot \cfrac{{{{\left( {t - 1} \right)}^{ - \cfrac{1}{2} + 1}}}}{{ - \cfrac{1}{2} + 1}} + C = - \cfrac{2}{a}\sqrt {t - 1} + C$

$= - \cfrac{2}{a}\sqrt {\cfrac{a}{x} - 1} + C = - \cfrac{2}{a}\sqrt {\cfrac{{a - x}}{x}} + C$