Integral de $$$\ln\left(x y\right)$$$ con respecto a $$$x$$$

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

Sea $$$u=x y$$$.

Entonces $$$du=\left(x y\right)^{\prime }dx = y dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{y}$$$.

Entonces,

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

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

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

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

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

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

Por lo tanto,

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

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

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

Recordemos que $$$u=x y$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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