Задача №1682

\[ \int\limits_{\sqrt{2}}^{+\infty}\frac{dx}{x\sqrt{x^2-1}} =\lim_{b\to+\infty}\int\limits_{\sqrt{2}}^{b}\frac{dx}{x\sqrt{x^2-1}} =\lim_{b\to+\infty}\int\limits_{\sqrt{2}}^{b}\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\\ =\lim_{b\to+\infty}\int\limits_{\sqrt{2}}^{b}\frac{-d\left(\frac{1}{x}\right)}{\sqrt{1-\left(\frac{1}{x}\right)^2}} =\lim_{b\to+\infty}\left(-\left. \arcsin\frac{1}{x}\right|_{\sqrt{2}}^{b}\right) =\lim_{b\to+\infty}\left(\arcsin\frac{1}{\sqrt{2}}-\arcsin\frac{1}{b}\right) =\frac{\pi}{4}. \]