Задача №1559

\[ \int\frac{x^4dx}{\left(1+x^2\right)^3} =\left[\begin{aligned} & u=x^3;\;du=3x^2dx;\\ & dv=\frac{xdx}{\left(1+x^2\right)^3};\;v=-\frac{1}{4\left(1+x^2\right)^2}. \end{aligned}\right] =-\frac{x^3}{4\left(1+x^2\right)}+\frac{3}{4}\int\frac{x^2dx}{\left(1+x^2\right)^2}=\\ =\left[\begin{aligned} & u=x;\;du=dx;\\ & dv=\frac{xdx}{\left(1+x^2\right)^2};\;v=-\frac{1}{2\left(1+x^2\right)}. \end{aligned}\right] =-\frac{x^3}{4\left(1+x^2\right)}-\frac{3}{8}\cdot\frac{x}{1+x^2}+\frac{3}{8}\int\frac{dx}{1+x^2}=\\ =-\frac{x^3}{4\left(1+x^2\right)}-\frac{3}{8}\cdot\frac{x}{1+x^2}+\frac{3}{8}\arctg{x}+C. \]