Integral de $$$- 8 a l t \left(t - 1\right) e^{- 5 t}$$$ em relação a $$$t$$$

Encontre $$$\int \left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)\, dt$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=- 8 a l$$$ e $$$f{\left(t \right)} = t \left(t - 1\right) e^{- 5 t}$$$:

$${\color{red}{\int{\left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)d t}}} = {\color{red}{\left(- 8 a l \int{t \left(t - 1\right) e^{- 5 t} d t}\right)}}$$

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

Sejam $$$\operatorname{u}=t \left(t - 1\right)$$$ e $$$\operatorname{dv}=e^{- 5 t} dt$$$.

Então $$$\operatorname{du}=\left(t \left(t - 1\right)\right)^{\prime }dt=\left(2 t - 1\right) dt$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 5 t} d t}=- \frac{e^{- 5 t}}{5}$$$ (os passos podem ser vistos »).

Assim,

$$- 8 a l {\color{red}{\int{t \left(t - 1\right) e^{- 5 t} d t}}}=- 8 a l {\color{red}{\left(t \left(t - 1\right) \cdot \left(- \frac{e^{- 5 t}}{5}\right)-\int{\left(- \frac{e^{- 5 t}}{5}\right) \cdot \left(2 t - 1\right) d t}\right)}}=- 8 a l {\color{red}{\left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - \int{\frac{\left(1 - 2 t\right) e^{- 5 t}}{5} d t}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=\frac{1}{5}$$$ e $$$f{\left(t \right)} = \left(1 - 2 t\right) e^{- 5 t}$$$:

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - {\color{red}{\int{\frac{\left(1 - 2 t\right) e^{- 5 t}}{5} d t}}}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - {\color{red}{\left(\frac{\int{\left(1 - 2 t\right) e^{- 5 t} d t}}{5}\right)}}\right)$$

Para a integral $$$\int{\left(1 - 2 t\right) e^{- 5 t} d t}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=1 - 2 t$$$ e $$$\operatorname{dv}=e^{- 5 t} dt$$$.

Então $$$\operatorname{du}=\left(1 - 2 t\right)^{\prime }dt=- 2 dt$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 5 t} d t}=- \frac{e^{- 5 t}}{5}$$$ (os passos podem ser vistos »).

A integral pode ser reescrita como

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - \frac{{\color{red}{\int{\left(1 - 2 t\right) e^{- 5 t} d t}}}}{5}\right)=- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - \frac{{\color{red}{\left(\left(1 - 2 t\right) \cdot \left(- \frac{e^{- 5 t}}{5}\right)-\int{\left(- \frac{e^{- 5 t}}{5}\right) \cdot \left(-2\right) d t}\right)}}}{5}\right)=- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} - \frac{{\color{red}{\left(- \frac{\left(1 - 2 t\right) e^{- 5 t}}{5} - \int{\frac{2 e^{- 5 t}}{5} d t}\right)}}}{5}\right)$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=\frac{2}{5}$$$ e $$$f{\left(t \right)} = e^{- 5 t}$$$:

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{{\color{red}{\int{\frac{2 e^{- 5 t}}{5} d t}}}}{5}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{{\color{red}{\left(\frac{2 \int{e^{- 5 t} d t}}{5}\right)}}}{5}\right)$$

Seja $$$u=- 5 t$$$.

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

A integral pode ser reescrita como

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{2 {\color{red}{\int{e^{- 5 t} d t}}}}{25}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{25}\right)$$

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

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{25}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} + \frac{2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{5}\right)}}}{25}\right)$$

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

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{125}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} - \frac{2 {\color{red}{e^{u}}}}{125}\right)$$

Recorde que $$$u=- 5 t$$$:

$$- 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} - \frac{2 e^{{\color{red}{u}}}}{125}\right) = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} - \frac{2 e^{{\color{red}{\left(- 5 t\right)}}}}{125}\right)$$

Portanto,

$$\int{\left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)d t} = - 8 a l \left(- \frac{t \left(t - 1\right) e^{- 5 t}}{5} + \frac{\left(1 - 2 t\right) e^{- 5 t}}{25} - \frac{2 e^{- 5 t}}{125}\right)$$

Simplifique:

$$\int{\left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)d t} = \frac{8 a l \left(25 t^{2} - 15 t - 3\right) e^{- 5 t}}{125}$$

Adicione a constante de integração:

$$\int{\left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)d t} = \frac{8 a l \left(25 t^{2} - 15 t - 3\right) e^{- 5 t}}{125}+C$$

$$$\int \left(- 8 a l t \left(t - 1\right) e^{- 5 t}\right)\, dt = \frac{8 a l \left(25 t^{2} - 15 t - 3\right) e^{- 5 t}}{125} + C$$$A