Задача №1512

\[ \int\sqrt{\frac{a-x}{x-b}}dx =\left[t=\sqrt{\frac{a-x}{x-b}};\;x=\frac{a+bt^2}{1+t^2};\;dx=\frac{2(b-a)tdt}{\left(1+t^2\right)^2}\right]=\\ =2(b-a)\cdot\int\frac{t^2dt}{\left(1+t^2\right)^2}= \left[\begin{aligned} & u=t;\;du=dt.\\ & dv=\frac{tdt}{\left(1+t^2\right)^2};\;v=-\frac{1}{2\left(1+t^2\right)}. \end{aligned}\right]=\\ =-(b-a)\frac{t}{1+t^2}+(b-a)\int\frac{dt}{1+t^2} =\sqrt{(a-x)(x-b)}-(a-b)\arctg\sqrt{\frac{a-x}{x-b}}+C. \]