Задача №1539

\[ \int{e^{-x^2}}x^5dx =[t=x^2] =\frac{1}{2}\int{t^2e^{-t}}dt =\left[\begin{aligned} & u=t^2;\;du=2tdt;\\ & dv=e^{-t}dt;\;v=-e^{-t}. \end{aligned}\right] =-\frac{t^2e^{-t}}{2}+\int{te^{-t}}dt=\\ =\left[\begin{aligned} & u=t;\;du=dt;\\ & dv=e^{-t}dt;\;v=-e^{-t}. \end{aligned}\right] =-\frac{t^2e^{-t}}{2}-te^{-t}+\int{e^{-t}}dt=\\ =-\frac{t^2e^{-t}}{2}-te^{-t}-e^{-t} =e^{-x^2}\cdot\left(-\frac{x^4}{2}-x^2-1\right)+C \]