$\int[e^x\sec x+e^x.\log(\sec x+\tan x)]dx$ is:

$-e^x\log(\sec x-\tan x)+C$

$\int[e^x\sec x+e^x.\log(\sec x+\tan x)]dx=\int e^x[\log (\sec x+\tan x)+\sec x]dx$

$\int e^x[\log (\sec x+\tan x)+\frac{d}{dx}[\log(\sec x+\tan x)]]dx=e^x.\log(\sec x+\tan x)+c$

$e^x\log(\sec x+\tan x)+C$

$=-e^x\log\left(\frac{1}{\sec x+\tan x}×\frac{\sec x-\tan x}{\sec x-\tan x}\right)+C$

$=-e^x\left(\frac{\sec x-\tan x}{\sec^2 x-\tan^2 x}\right)+C$

$=-e^x.\log(\sec x-\tan x)+c$