Integral de $$$x^{5} e^{- x^{2}}$$$

Encontre $$$\int x^{5} e^{- x^{2}}\, dx$$$.

Seja $$$u=- x^{2}$$$.

Então $$$du=\left(- x^{2}\right)^{\prime }dx = - 2 x dx$$$ (veja os passos »), e obtemos $$$x dx = - \frac{du}{2}$$$.

A integral torna-se

$${\color{red}{\int{x^{5} e^{- x^{2}} d x}}} = {\color{red}{\int{\left(- \frac{u^{2} e^{u}}{2}\right)d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{2}$$$ e $$$f{\left(u \right)} = u^{2} e^{u}$$$:

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

Para a integral $$$\int{u^{2} e^{u} d u}$$$, use integração por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Sejam $$$\operatorname{g}=u^{2}$$$ e $$$\operatorname{dv}=e^{u} du$$$.

Então $$$\operatorname{dg}=\left(u^{2}\right)^{\prime }du=2 u du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=2$$$ e $$$f{\left(u \right)} = u e^{u}$$$:

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

Para a integral $$$\int{u e^{u} d u}$$$, use integração por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Sejam $$$\operatorname{g}=u$$$ e $$$\operatorname{dv}=e^{u} du$$$.

Então $$$\operatorname{dg}=\left(u\right)^{\prime }du=1 du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (os passos podem ser vistos »).

Assim,

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

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

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

Recorde que $$$u=- x^{2}$$$:

$$- e^{{\color{red}{u}}} + {\color{red}{u}} e^{{\color{red}{u}}} - \frac{{\color{red}{u}}^{2} e^{{\color{red}{u}}}}{2} = - e^{{\color{red}{\left(- x^{2}\right)}}} + {\color{red}{\left(- x^{2}\right)}} e^{{\color{red}{\left(- x^{2}\right)}}} - \frac{{\color{red}{\left(- x^{2}\right)}}^{2} e^{{\color{red}{\left(- x^{2}\right)}}}}{2}$$

Portanto,

$$\int{x^{5} e^{- x^{2}} d x} = - \frac{x^{4} e^{- x^{2}}}{2} - x^{2} e^{- x^{2}} - e^{- x^{2}}$$

Simplifique:

$$\int{x^{5} e^{- x^{2}} d x} = \left(- \frac{x^{4}}{2} - x^{2} - 1\right) e^{- x^{2}}$$

Adicione a constante de integração:

$$\int{x^{5} e^{- x^{2}} d x} = \left(- \frac{x^{4}}{2} - x^{2} - 1\right) e^{- x^{2}}+C$$

$$$\int x^{5} e^{- x^{2}}\, dx = \left(- \frac{x^{4}}{2} - x^{2} - 1\right) e^{- x^{2}} + C$$$A