Integral de $$$\ln\left(- 5 x\right)$$$

Halla $$$\int \ln\left(- 5 x\right)\, dx$$$.

Sea $$$u=- 5 x$$$.

Entonces $$$du=\left(- 5 x\right)^{\prime }dx = - 5 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - \frac{du}{5}$$$.

Por lo tanto,

$${\color{red}{\int{\ln{\left(- 5 x \right)} d x}}} = {\color{red}{\int{\left(- \frac{\ln{\left(u \right)}}{5}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{5}$$$ y $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

$${\color{red}{\int{\left(- \frac{\ln{\left(u \right)}}{5}\right)d u}}} = {\color{red}{\left(- \frac{\int{\ln{\left(u \right)} d u}}{5}\right)}}$$

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,

$$- \frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{5}=- \frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{5}=- \frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{5}$$

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

$$- \frac{u \ln{\left(u \right)}}{5} + \frac{{\color{red}{\int{1 d u}}}}{5} = - \frac{u \ln{\left(u \right)}}{5} + \frac{{\color{red}{u}}}{5}$$

Recordemos que $$$u=- 5 x$$$:

$$\frac{{\color{red}{u}}}{5} - \frac{{\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{5} = \frac{{\color{red}{\left(- 5 x\right)}}}{5} - \frac{{\color{red}{\left(- 5 x\right)}} \ln{\left({\color{red}{\left(- 5 x\right)}} \right)}}{5}$$

Por lo tanto,

$$\int{\ln{\left(- 5 x \right)} d x} = x \ln{\left(- 5 x \right)} - x$$

Simplificar:

$$\int{\ln{\left(- 5 x \right)} d x} = x \left(\ln{\left(- x \right)} - 1 + \ln{\left(5 \right)}\right)$$

Añade la constante de integración:

$$\int{\ln{\left(- 5 x \right)} d x} = x \left(\ln{\left(- x \right)} - 1 + \ln{\left(5 \right)}\right)+C$$

$$$\int \ln\left(- 5 x\right)\, dx = x \left(\ln\left(- x\right) - 1 + \ln\left(5\right)\right) + C$$$A