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

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

Integrate term by term:

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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