$\begin{aligned} & \int_2^5 \sqrt{\frac{5-x}{x-2}} d x=\int_2^5 \sqrt{\frac{(5-x)^2}{(x-2)(5-x)}} d x \\ = & \int_2^5 \frac{5-x}{\sqrt{7 x-x^2-10}} d x .\end{aligned}$
$\begin{aligned} & \therefore \quad 5-x=A \frac{d}{d x}\left(7 x-x^2-10\right)+B \\ & =(7-2 x) A+B \\ & -2 A=-1,7 A+B=5 \\ & A=\frac{1}{2} ; \quad B=5-\frac{7}{2}=\frac{3}{2} \\ & \therefore \quad 5-x=\frac{1}{2} \frac{d}{d x}\left(7 x-x^2-10\right)+\frac{3}{2} \\ & =\frac{1}{2}(7-2 x)+\frac{3}{2} \\ & \therefore \quad \int_2^5 \frac{5-x}{\sqrt{7 x-x^2-10}} d x \\ & =\frac{1}{2} \int_2^5 \frac{7-2 x}{\sqrt{7 x-x^2-10}} d x+\frac{3}{2} \int_2^5 \frac{d x}{\sqrt{7 x-x^2-10}} \\ & =\frac{1}{2} \int_2^5 \frac{7-2 x}{\sqrt{7 x-x^2-10} d x+\frac{3}{2} \int_2^5 \frac{d x}{\sqrt{\left(\frac{3}{2}\right)^2-\left(x-\frac{7}{2}\right)^2}}} \\ & \therefore \quad \\ & =\frac{1}{2} \cdot 2\left[\sqrt{7 x-x^2-10}\right]_2^5+\frac{3}{2}\left[\sin ^{-1} \frac{x-\frac{7}{2}}{\frac{3}{2}}\right]_2^5\end{aligned}$
$\begin{aligned} & =0+\frac{3}{2}\left[\sin ^{-1}\left(\frac{2 x-7}{3}\right)\right]_2^5 \\ & =\frac{3}{2}\left[\sin ^{-1} 1-\sin ^{-1}(-1)\right]=\frac{3}{2}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)=\frac{3 \pi}{2} .\end{aligned}$