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

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

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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$- \int{\frac{1}{x - 1} d x} - {\color{red}{\int{1 d x}}} = - \int{\frac{1}{x - 1} d x} - {\color{red}{x}}$$

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

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

So,

$$- x - {\color{red}{\int{\frac{1}{x - 1} d x}}} = - x - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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