Задача №1396

\[ \int\arctg\sqrt{x}dx =\left[\begin{aligned}& u=\arctg{\sqrt{x}};\,du=\frac{dx}{2\sqrt{x}(1+x)}.\\& dv=dx;\,v=x.\end{aligned}\right] =x\arctg\sqrt{x}-\frac{1}{2}\int\frac{xdx}{\sqrt{x}(1+x)}=\\ =x\arctg\sqrt{x}-\int\frac{xd(\sqrt{x})}{x+1} =x\arctg\sqrt{x}-\int\frac{x+1-1}{x+1}d(\sqrt{x})=\\ =x\arctg\sqrt{x}-\int\left(1-\frac{1}{1+\left(\sqrt{x}\right)^2}\right)d(\sqrt{x}) =x\arctg\sqrt{x}-\sqrt{x}+\arctg\sqrt{x}+C =(x+1)\arctg\sqrt{x}-\sqrt{x}+C \]