Integral de $$$\ln\left(f x\right)$$$ em relação a $$$x$$$

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

Seja $$$u=f x$$$.

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

Logo,

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

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

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

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

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

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

A integral torna-se

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

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

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

Recorde que $$$u=f x$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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