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

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

Para a integral $$$\int{x^{2} e^{- 3 x} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=x^{2}$$$ e $$$\operatorname{dv}=e^{- 3 x} dx$$$.

Então $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 3 x} d x}=- \frac{e^{- 3 x}}{3}$$$ (os passos podem ser vistos »).

Portanto,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=- \frac{2}{3}$$$ e $$$f{\left(x \right)} = x e^{- 3 x}$$$:

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

Para a integral $$$\int{x e^{- 3 x} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=x$$$ e $$$\operatorname{dv}=e^{- 3 x} dx$$$.

Então $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 3 x} d x}=- \frac{e^{- 3 x}}{3}$$$ (os passos podem ser vistos »).

Logo,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=- \frac{1}{3}$$$ e $$$f{\left(x \right)} = e^{- 3 x}$$$:

$$- \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 {\color{red}{\int{\left(- \frac{e^{- 3 x}}{3}\right)d x}}}}{3} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 {\color{red}{\left(- \frac{\int{e^{- 3 x} d x}}{3}\right)}}}{3}$$

Seja $$$u=- 3 x$$$.

Então $$$du=\left(- 3 x\right)^{\prime }dx = - 3 dx$$$ (veja os passos »), e obtemos $$$dx = - \frac{du}{3}$$$.

Assim,

$$- \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} + \frac{2 {\color{red}{\int{e^{- 3 x} d x}}}}{9} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}}}{9}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{3}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$$- \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}}}{9} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} + \frac{2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{3}\right)}}}{9}$$

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

$$- \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{27} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 {\color{red}{e^{u}}}}{27}$$

Recorde que $$$u=- 3 x$$$:

$$- \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 e^{{\color{red}{u}}}}{27} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 e^{{\color{red}{\left(- 3 x\right)}}}}{27}$$

Portanto,

$$\int{x^{2} e^{- 3 x} d x} = - \frac{x^{2} e^{- 3 x}}{3} - \frac{2 x e^{- 3 x}}{9} - \frac{2 e^{- 3 x}}{27}$$

Simplifique:

$$\int{x^{2} e^{- 3 x} d x} = \frac{\left(- 9 x^{2} - 6 x - 2\right) e^{- 3 x}}{27}$$

Adicione a constante de integração:

$$\int{x^{2} e^{- 3 x} d x} = \frac{\left(- 9 x^{2} - 6 x - 2\right) e^{- 3 x}}{27}+C$$

$$$\int x^{2} e^{- 3 x}\, dx = \frac{\left(- 9 x^{2} - 6 x - 2\right) e^{- 3 x}}{27} + C$$$A