Integral de $$$\frac{\left(- \frac{10 - x}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5}$$$

Encontre $$$\int \frac{\left(- \frac{10 - x}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5}\, dx$$$.

A entrada é reescrita como: $$$\int{\frac{\left(- \frac{10 - x}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5} d x}=\int{\left(\frac{x - 10}{5 e^{\frac{1}{10}}} + \frac{1}{5}\right) e^{- \frac{x}{5}} d x}$$$.

Simplifique o integrando:

$${\color{red}{\int{\left(\frac{x - 10}{5 e^{\frac{1}{10}}} + \frac{1}{5}\right) e^{- \frac{x}{5}} d x}}} = {\color{red}{\int{\frac{\left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5} d x}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{5}$$$ e $$$f{\left(x \right)} = \left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}$$$:

$${\color{red}{\int{\frac{\left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5} d x}}} = {\color{red}{\left(\frac{\int{\left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}} d x}}{5}\right)}}$$

Para a integral $$$\int{\left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}$$$ e $$$\operatorname{dv}=e^{- \frac{x}{5}} dx$$$.

Então $$$\operatorname{du}=\left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}}\right)^{\prime }dx=\frac{dx}{e^{\frac{1}{10}}}$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- \frac{x}{5}} d x}=- 5 e^{- \frac{x}{5}}$$$ (os passos podem ser vistos »).

A integral pode ser reescrita como

$$\frac{{\color{red}{\int{\left(\frac{x - 10}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}} d x}}}}{5}=\frac{{\color{red}{\left(\frac{x - 10 + e^{\frac{1}{10}}}{e^{\frac{1}{10}}} \cdot \left(- 5 e^{- \frac{x}{5}}\right)-\int{\left(- 5 e^{- \frac{x}{5}}\right) \cdot e^{- \frac{1}{10}} d x}\right)}}}{5}=\frac{{\color{red}{\left(- \frac{5 \left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \int{\left(- \frac{5 e^{- \frac{x}{5}}}{e^{\frac{1}{10}}}\right)d x}\right)}}}{5}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=- \frac{5}{e^{\frac{1}{10}}}$$$ e $$$f{\left(x \right)} = e^{- \frac{x}{5}}$$$:

$$- \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{{\color{red}{\int{\left(- \frac{5 e^{- \frac{x}{5}}}{e^{\frac{1}{10}}}\right)d x}}}}{5} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{{\color{red}{\left(- \frac{5 \int{e^{- \frac{x}{5}} d x}}{e^{\frac{1}{10}}}\right)}}}{5}$$

Seja $$$u=- \frac{x}{5}$$$.

Então $$$du=\left(- \frac{x}{5}\right)^{\prime }dx = - \frac{dx}{5}$$$ (veja os passos »), e obtemos $$$dx = - 5 du$$$.

A integral pode ser reescrita como

$$- \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} + \frac{{\color{red}{\int{e^{- \frac{x}{5}} d x}}}}{e^{\frac{1}{10}}} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} + \frac{{\color{red}{\int{\left(- 5 e^{u}\right)d u}}}}{e^{\frac{1}{10}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-5$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$$- \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} + \frac{{\color{red}{\int{\left(- 5 e^{u}\right)d u}}}}{e^{\frac{1}{10}}} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} + \frac{{\color{red}{\left(- 5 \int{e^{u} d u}\right)}}}{e^{\frac{1}{10}}}$$

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

$$- \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{5 {\color{red}{\int{e^{u} d u}}}}{e^{\frac{1}{10}}} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{5 {\color{red}{e^{u}}}}{e^{\frac{1}{10}}}$$

Recorde que $$$u=- \frac{x}{5}$$$:

$$- \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{5 e^{{\color{red}{u}}}}{e^{\frac{1}{10}}} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{5 e^{{\color{red}{\left(- \frac{x}{5}\right)}}}}{e^{\frac{1}{10}}}$$

Portanto,

$$\int{\left(\frac{x - 10}{5 e^{\frac{1}{10}}} + \frac{1}{5}\right) e^{- \frac{x}{5}} d x} = - \frac{\left(x - 10 + e^{\frac{1}{10}}\right) e^{- \frac{x}{5}}}{e^{\frac{1}{10}}} - \frac{5 e^{- \frac{x}{5}}}{e^{\frac{1}{10}}}$$

Simplifique:

$$\int{\left(\frac{x - 10}{5 e^{\frac{1}{10}}} + \frac{1}{5}\right) e^{- \frac{x}{5}} d x} = \left(- x - e^{\frac{1}{10}} + 5\right) e^{- \frac{x}{5} - \frac{1}{10}}$$

Adicione a constante de integração:

$$\int{\left(\frac{x - 10}{5 e^{\frac{1}{10}}} + \frac{1}{5}\right) e^{- \frac{x}{5}} d x} = \left(- x - e^{\frac{1}{10}} + 5\right) e^{- \frac{x}{5} - \frac{1}{10}}+C$$

$$$\int \frac{\left(- \frac{10 - x}{e^{\frac{1}{10}}} + 1\right) e^{- \frac{x}{5}}}{5}\, dx = \left(- x - e^{\frac{1}{10}} + 5\right) e^{- \frac{x}{5} - \frac{1}{10}} + C$$$A