$\int \sec x·\log \left(\sec x+\tan x\right)dx$

$$\begin{gathered} \int \sec x·\log \left(\sec x+\tan x\right)dx \\ \mathrm{Let}\log \left(\sec x+\tan x\right)=t \\ \Rightarrow \frac{\left(\sec x\tan x+{\sec }^{2}x\right)}{\left(\sec x+\tan x\right)}=\frac{dt}{dx} \\ \Rightarrow \frac{\sec x\left(\sec x+\tan x\right)}{\left(\sec x+\tan x\right)}dx=dt \\ \mathrm{Now},\int \sec x·\log \left(\sec x+\tan x\right)dx \\ =\int t·dt \\ =\frac{t^2}{2}+\mathrm{C} \\ =\frac{{\left[\log \left(\sec x+\tan x\right)\right]}^2}{2}+\mathrm{C} \end{gathered}$$