Задача №1394

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