Q.Evaluate: $\int \sqrt{\dfrac{9-x}{x}}\,dx$

Multiply numerator and denominator inside the root by $\sqrt{9-x}$: $\sqrt{\dfrac{9-x}{x}}=\dfrac{9-x}{\sqrt{x(9-x)}}=\dfrac{9-x}{\sqrt{9x-x^2}}$.

Split: $\displaystyle9\int\dfrac{dx}{\sqrt{9x-x^2}}-\int\dfrac{x\,dx}{\sqrt{9x-x^2}}$.

For the second piece, write $x=-\dfrac12(9-2x)+\dfrac92$ (since $\dfrac{d}{dx}(9x-x^2)=9-2x$): $\displaystyle\int\dfrac{x\,dx}{\sqrt{9x-x^2}}=-\sqrt{9x-x^2}+\dfrac92\int\dfrac{dx}{\sqrt{9x-x^2}}$.

Combining, the two $\int\dfrac{dx}{\sqrt{9x-x^2}}$ pieces give net coefficient $9-\dfrac92=\dfrac92$, plus $+\sqrt{9x-x^2}$ from the sign flip. …

Rationalise the surd, split, use the derivative-of-radicand trick o …

Losing track of the sign when combining the two $\int dx/\sqrt{9x-x^2}$ contributions (one from the direc …