Integral of $-2 + \frac{1}{x}$

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

Integrate term by term:

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

Apply the constant rule $\int c\, dx = c x$ with $c=2$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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