Integral of $$$\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}$$$

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

Expand the expression:

$${\color{red}{\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(- \tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + \int{\sec^{2}{\left(x \right)} d x}\right)}}$$

The integral of $$$\sec^{2}{\left(x \right)}$$$ is $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$:

$$- \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + {\color{red}{\int{\sec^{2}{\left(x \right)} d x}}} = - \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + {\color{red}{\tan{\left(x \right)}}}$$

The integral of $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ is $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$:

$$\tan{\left(x \right)} - {\color{red}{\int{\tan{\left(x \right)} \sec{\left(x \right)} d x}}} = \tan{\left(x \right)} - {\color{red}{\sec{\left(x \right)}}}$$

Therefore,

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x} = \tan{\left(x \right)} - \sec{\left(x \right)}$$

Add the constant of integration:

$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x} = \tan{\left(x \right)} - \sec{\left(x \right)}+C$$

$$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}\, dx = \left(\tan{\left(x \right)} - \sec{\left(x \right)}\right) + C$$$A