Integral of $$$\frac{f_{1} \tan^{2}{\left(f \right)}}{g}$$$ with respect to $$$g$$$

Find $$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg$$$.

Apply the constant multiple rule $$$\int c f{\left(g \right)}\, dg = c \int f{\left(g \right)}\, dg$$$ with $$$c=f_{1} \tan^{2}{\left(f \right)}$$$ and $$$f{\left(g \right)} = \frac{1}{g}$$$:

$${\color{red}{\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g}}} = {\color{red}{f_{1} \tan^{2}{\left(f \right)} \int{\frac{1}{g} d g}}}$$

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

$$f_{1} \tan^{2}{\left(f \right)} {\color{red}{\int{\frac{1}{g} d g}}} = f_{1} \tan^{2}{\left(f \right)} {\color{red}{\ln{\left(\left|{g}\right| \right)}}}$$

Therefore,

$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}$$

Add the constant of integration:

$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}+C$$

$$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg = f_{1} \ln\left(\left|{g}\right|\right) \tan^{2}{\left(f \right)} + C$$$A