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

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

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

Then $$$du=\left(1 - \phi\right)^{\prime }d\phi = - d\phi$$$ (steps can be seen »), and we have that $$$d\phi = - du$$$.

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:

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

For the integral $$$\int{\ln{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{o} \operatorname{dv} = \operatorname{o}\operatorname{v} - \int \operatorname{v} \operatorname{do}$$$.

Let $$$\operatorname{o}=\ln{\left(u \right)}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{do}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

Thus,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Recall that $$$u=1 - \phi$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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