[m] ∫_{0}^{+ ∞} \frac{\sqrt{arc ctgx}}{1+x^2}dx=∫_{0}^{+ ∞} \sqrt{arc ctgx}\cdot (-d(arc ctgx))=-∫_{0}^{+ ∞}( arc ctgx)^{\frac{1}{2}}d (arc ctgx)=(\frac{( arcctgx)^{\frac{1}{2}+1}}{\frac{1}{2}+1})|_{0}^{+ ∞}=[/m]
[m]=-lim_{x → + ∞ }\frac{2(arc ctgx)^{\frac{3}{2}}}{3}+\frac{2(arc ctg 0)^{\frac{3}{2}}}{3}=-0+\frac{2(\frac{π}{2})^{\frac{3}{2}}}{3}=\frac{π\sqrt{\frac{π}{2}}}{3}[/m]