Integral de $$$\ln\left(12 x\right)$$$

Encontre $$$\int \ln\left(12 x\right)\, dx$$$.

Seja $$$u=12 x$$$.

Então $$$du=\left(12 x\right)^{\prime }dx = 12 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{12}$$$.

Logo,

$${\color{red}{\int{\ln{\left(12 x \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{12} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{12}$$$ e $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{12} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{12}\right)}}$$

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 »).

Assim,

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{12}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{12}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{12}$$

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

$$\frac{u \ln{\left(u \right)}}{12} - \frac{{\color{red}{\int{1 d u}}}}{12} = \frac{u \ln{\left(u \right)}}{12} - \frac{{\color{red}{u}}}{12}$$

Recorde que $$$u=12 x$$$:

$$- \frac{{\color{red}{u}}}{12} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{12} = - \frac{{\color{red}{\left(12 x\right)}}}{12} + \frac{{\color{red}{\left(12 x\right)}} \ln{\left({\color{red}{\left(12 x\right)}} \right)}}{12}$$

Portanto,

$$\int{\ln{\left(12 x \right)} d x} = x \ln{\left(12 x \right)} - x$$

Simplifique:

$$\int{\ln{\left(12 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(12 \right)}\right)$$

Adicione a constante de integração:

$$\int{\ln{\left(12 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(12 \right)}\right)+C$$

$$$\int \ln\left(12 x\right)\, dx = x \left(\ln\left(x\right) - 1 + \ln\left(12\right)\right) + C$$$A