Integral de $$$\ln\left(10 x\right)$$$

Encontre $$$\int \ln\left(10 x\right)\, dx$$$.

Seja $$$u=10 x$$$.

Então $$$du=\left(10 x\right)^{\prime }dx = 10 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{10}$$$.

Assim,

$${\color{red}{\int{\ln{\left(10 x \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{10} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{10}$$$ e $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{10} d u}}} = {\color{red}{\left(\frac{\int{\ln{\left(u \right)} d u}}{10}\right)}}$$

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

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

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

Logo,

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{10}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{10}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{10}$$

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

$$\frac{u \ln{\left(u \right)}}{10} - \frac{{\color{red}{\int{1 d u}}}}{10} = \frac{u \ln{\left(u \right)}}{10} - \frac{{\color{red}{u}}}{10}$$

Recorde que $$$u=10 x$$$:

$$- \frac{{\color{red}{u}}}{10} + \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{10} = - \frac{{\color{red}{\left(10 x\right)}}}{10} + \frac{{\color{red}{\left(10 x\right)}} \ln{\left({\color{red}{\left(10 x\right)}} \right)}}{10}$$

Portanto,

$$\int{\ln{\left(10 x \right)} d x} = x \ln{\left(10 x \right)} - x$$

Simplifique:

$$\int{\ln{\left(10 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(10 \right)}\right)$$

Adicione a constante de integração:

$$\int{\ln{\left(10 x \right)} d x} = x \left(\ln{\left(x \right)} - 1 + \ln{\left(10 \right)}\right)+C$$

$$$\int \ln\left(10 x\right)\, dx = x \left(\ln\left(x\right) - 1 + \ln\left(10\right)\right) + C$$$A