Integral de $$$\ln\left(4 - 2 x\right)$$$

Encontre $$$\int \ln\left(4 - 2 x\right)\, dx$$$.

Seja $$$u=4 - 2 x$$$.

Então $$$du=\left(4 - 2 x\right)^{\prime }dx = - 2 dx$$$ (veja os passos »), e obtemos $$$dx = - \frac{du}{2}$$$.

A integral torna-se

$${\color{red}{\int{\ln{\left(4 - 2 x \right)} d x}}} = {\color{red}{\int{\left(- \frac{\ln{\left(u \right)}}{2}\right)d u}}}$$

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 $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

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

Para a integral $$$\int{\ln{\left(u \right)} d u}$$$, use integração por partes $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.

Sejam $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ e $$$\operatorname{dv}=du$$$.

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

A integral pode ser reescrita como

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

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

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

Recorde que $$$u=4 - 2 x$$$:

$$\frac{{\color{red}{u}}}{2} - \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{2} = \frac{{\color{red}{\left(4 - 2 x\right)}}}{2} - \frac{{\color{red}{\left(4 - 2 x\right)}} \ln{\left({\color{red}{\left(4 - 2 x\right)}} \right)}}{2}$$

Portanto,

$$\int{\ln{\left(4 - 2 x \right)} d x} = - x - \frac{\left(4 - 2 x\right) \ln{\left(4 - 2 x \right)}}{2} + 2$$

Simplifique:

$$\int{\ln{\left(4 - 2 x \right)} d x} = \left(x - 2\right) \left(\ln{\left(2 - x \right)} - 1 + \ln{\left(2 \right)}\right)$$

Adicione a constante de integração:

$$\int{\ln{\left(4 - 2 x \right)} d x} = \left(x - 2\right) \left(\ln{\left(2 - x \right)} - 1 + \ln{\left(2 \right)}\right)+C$$

$$$\int \ln\left(4 - 2 x\right)\, dx = \left(x - 2\right) \left(\ln\left(2 - x\right) - 1 + \ln\left(2\right)\right) + C$$$A