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Question Number 42182 by mondodotto@gmail.com last updated on 19/Aug/18

$$\mathbf{integrate}\int {\mathbf{e}}^{{\mathit{x}}^2}{\mathit{x}}^2\mathit{d}\mathit{x}$$

Commented by mondodotto@gmail.com last updated on 19/Aug/18

$$\mathbf{please}\mathbf{help}\mathbf{this}$$

Commented by maxmathsup by imad last updated on 19/Aug/18

$$I=\int x{e}^{x^2}xdxandbyparts{u}^{{}^{\prime }}=x{e}^{x^2}andv=x\Rightarrow$$$$I=\frac{x}{2}{e}^{x^2}-\int \frac{1}{2}{e}^{x^2}dx=\frac{x}{2}{e}^{x^2}-\frac{1}{2}\int e^{x^2}dxbut$$$$e^{x^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{n!}\Rightarrow \int e^{x^2}dx?=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!(2n+1)}{x}^{2n+1}\Rightarrow$$$$\int x^2{e}^{x^2}dx=\frac{x{e}^{x^2}}{2}-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{(2n+1)!}+c.$$

Commented by maxmathsup by imad last updated on 19/Aug/18

$$\int x^2{e}^{x^2}dx=\frac{x{e}^{x^2}}{2}-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{n!(2n+1)}+cletfindtheradiusofthisserie$$$$forx\ne 0\mid \frac{u_{n+1}\left(x\right)}{u_n\left(x\right)}\mid =\mid \frac{x^{2n+3}}{(n+1)!(2n+3)}\frac{n!(2n+1)}{x^{2n+1}}\mid =\frac{2n+1}{(2n+3)(n+1)}\mid x{\mid }^2\to 0(n\to +\mathrm{\infty })$$$$soR=+\mathrm{\infty }.$$

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