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

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dy = c y$$$ with $$$c=1$$$:

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

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

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

Therefore,

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

Add the constant of integration:

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

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