Integral de $$$- 4 x e^{- x}$$$

Encontre $$$\int \left(- 4 x e^{- x}\right)\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-4$$$ e $$$f{\left(x \right)} = x e^{- x}$$$:

$${\color{red}{\int{\left(- 4 x e^{- x}\right)d x}}} = {\color{red}{\left(- 4 \int{x e^{- x} d x}\right)}}$$

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

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

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

Logo,

$$- 4 {\color{red}{\int{x e^{- x} d x}}}=- 4 {\color{red}{\left(x \cdot \left(- e^{- x}\right)-\int{\left(- e^{- x}\right) \cdot 1 d x}\right)}}=- 4 {\color{red}{\left(- x e^{- x} - \int{\left(- e^{- x}\right)d x}\right)}}$$

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

$$4 x e^{- x} + 4 {\color{red}{\int{\left(- e^{- x}\right)d x}}} = 4 x e^{- x} + 4 {\color{red}{\left(- \int{e^{- x} d x}\right)}}$$

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

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

Portanto,

$$4 x e^{- x} - 4 {\color{red}{\int{e^{- x} d x}}} = 4 x e^{- x} - 4 {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

$$4 x e^{- x} - 4 {\color{red}{\int{\left(- e^{u}\right)d u}}} = 4 x e^{- x} - 4 {\color{red}{\left(- \int{e^{u} d u}\right)}}$$

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

$$4 x e^{- x} + 4 {\color{red}{\int{e^{u} d u}}} = 4 x e^{- x} + 4 {\color{red}{e^{u}}}$$

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

$$4 x e^{- x} + 4 e^{{\color{red}{u}}} = 4 x e^{- x} + 4 e^{{\color{red}{\left(- x\right)}}}$$

Portanto,

$$\int{\left(- 4 x e^{- x}\right)d x} = 4 x e^{- x} + 4 e^{- x}$$

Simplifique:

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

Adicione a constante de integração:

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

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