Integral of $$$- \frac{1}{1 - x}$$$

Find $$$\int \left(- \frac{1}{1 - x}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-1$$$ and $$$f{\left(x \right)} = \frac{1}{1 - x}$$$:

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

Let $$$u=1 - x$$$.

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

The integral becomes

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

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

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recall that $$$u=1 - x$$$:

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

Therefore,

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

Add the constant of integration:

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

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