Integral de $$$\ln\left(1 - \phi\right)$$$

Halla $$$\int \ln\left(1 - \phi\right)\, d\phi$$$.

Sea $$$u=1 - \phi$$$.

Entonces $$$du=\left(1 - \phi\right)^{\prime }d\phi = - d\phi$$$ (los pasos pueden verse »), y obtenemos que $$$d\phi = - du$$$.

Entonces,

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

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

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

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 »).

La integral puede reescribirse como

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

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

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

Recordemos que $$$u=1 - \phi$$$:

$${\color{red}{u}} - {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = {\color{red}{\left(1 - \phi\right)}} - {\color{red}{\left(1 - \phi\right)}} \ln{\left({\color{red}{\left(1 - \phi\right)}} \right)}$$

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

$$$\int \ln\left(1 - \phi\right)\, d\phi = \left(\phi - 1\right) \left(\ln\left(1 - \phi\right) - 1\right) + C$$$A