Integral of $$$\sin{\left(x \right)} - \frac{1}{x}$$$

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

Integrate term by term:

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

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

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

The integral of the sine is $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

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

Therefore,

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

Add the constant of integration:

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

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