Задача №1416

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