Integral de $$$x^{3} e^{x^{2}}$$$

Halla $$$\int x^{3} e^{x^{2}}\, dx$$$.

Sea $$$u=x^{2}$$$.

Entonces $$$du=\left(x^{2}\right)^{\prime }dx = 2 x dx$$$ (los pasos pueden verse »), y obtenemos que $$$x dx = \frac{du}{2}$$$.

Por lo tanto,

$${\color{red}{\int{x^{3} e^{x^{2}} d x}}} = {\color{red}{\int{\frac{u e^{u}}{2} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = u e^{u}$$$:

$${\color{red}{\int{\frac{u e^{u}}{2} d u}}} = {\color{red}{\left(\frac{\int{u e^{u} d u}}{2}\right)}}$$

Para la integral $$$\int{u e^{u} d u}$$$, utiliza la integración por partes $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

Sean $$$\operatorname{m}=u$$$ y $$$\operatorname{dv}=e^{u} du$$$.

Entonces $$$\operatorname{dm}=\left(u\right)^{\prime }du=1 du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (los pasos pueden verse »).

Por lo tanto,

$$\frac{{\color{red}{\int{u e^{u} d u}}}}{2}=\frac{{\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}}{2}=\frac{{\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}}{2}$$

La integral de la función exponencial es $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{u e^{u}}{2} - \frac{{\color{red}{\int{e^{u} d u}}}}{2} = \frac{u e^{u}}{2} - \frac{{\color{red}{e^{u}}}}{2}$$

Recordemos que $$$u=x^{2}$$$:

$$- \frac{e^{{\color{red}{u}}}}{2} + \frac{{\color{red}{u}} e^{{\color{red}{u}}}}{2} = - \frac{e^{{\color{red}{x^{2}}}}}{2} + \frac{{\color{red}{x^{2}}} e^{{\color{red}{x^{2}}}}}{2}$$

Por lo tanto,

$$\int{x^{3} e^{x^{2}} d x} = \frac{x^{2} e^{x^{2}}}{2} - \frac{e^{x^{2}}}{2}$$

Simplificar:

$$\int{x^{3} e^{x^{2}} d x} = \frac{\left(x^{2} - 1\right) e^{x^{2}}}{2}$$

Añade la constante de integración:

$$\int{x^{3} e^{x^{2}} d x} = \frac{\left(x^{2} - 1\right) e^{x^{2}}}{2}+C$$

$$$\int x^{3} e^{x^{2}}\, dx = \frac{\left(x^{2} - 1\right) e^{x^{2}}}{2} + C$$$A