Integral of $$$- 3 x_{2} + \frac{1}{x}$$$ with respect to $$$x$$$

Find $$$\int \left(- 3 x_{2} + \frac{1}{x}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{x} d x} - \int{3 x_{2} d x}\right)}}$$

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

$$- \int{3 x_{2} d x} + {\color{red}{\int{\frac{1}{x} d x}}} = - \int{3 x_{2} d x} + {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=3 x_{2}$$$:

$$\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{3 x_{2} d x}}} = \ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(3 x x_{2}\right)}}$$

Therefore,

$$\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x} = - 3 x x_{2} + \ln{\left(\left|{x}\right| \right)}$$

Add the constant of integration:

$$\int{\left(- 3 x_{2} + \frac{1}{x}\right)d x} = - 3 x x_{2} + \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(- 3 x_{2} + \frac{1}{x}\right)\, dx = \left(- 3 x x_{2} + \ln\left(\left|{x}\right|\right)\right) + C$$$A