Задача №1459

\[ \int\sqrt{\frac{x-1}{x+1}}\frac{dx}{x^2} =-\int\sqrt{\frac{1-\frac{1}{x}}{1+\frac{1}{x}}}d\left(\frac{1}{x}\right) =\left|z=\frac{1}{x}\right| =-\int\sqrt{\frac{1-z}{1+z}}dz =-\int\sqrt{\frac{(1-z)^2}{(1+z)(1-z)}}dz=\\ =-\int\frac{(1-z)dz}{\sqrt{1-z^2}} =-\int\frac{dz}{\sqrt{1-z^2}}+\int\frac{zdz}{\sqrt{1-z^2}} =-\int\frac{dz}{\sqrt{1-z^2}}-\frac{1}{2}\int\left(1-z^2\right)^{-\frac{1}{2}}d\left(1-z^2\right)=\\ =-\arcsin{z}-\left(1-z^2\right)^{\frac{1}{2}}+C =-\arcsin{z}-\sqrt{1-z^2}+C =-\arcsin\frac{1}{x}-\sqrt{1-\frac{1}{x^2}}+C \]