Integral de $$$\ln\left(\frac{1}{x}\right)$$$

Halla $$$\int \left(- \ln\left(x\right)\right)\, dx$$$.

La entrada se reescribe: $$$\int{\ln{\left(\frac{1}{x} \right)} d x}=\int{\left(- \ln{\left(x \right)}\right)d x}$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-1$$$ y $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:

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

Para la integral $$$\int{\ln{\left(x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(x \right)}$$$ y $$$\operatorname{dv}=dx$$$.

Entonces $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d x}=x$$$ (los pasos pueden verse »).

Entonces,

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

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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