$\int e^x \sec x(1+\tan x) d x$ equals :

$e^x \sec x+c$ $e^x \tan x+c$ $e^x \sin x+c$ $e^x \cos x+c$

$e^x \sec x+c$

$\int e^x \sec x(1+\tan x) d x$ $= \int e^x \sec x+e^x \sec x \tan x d x$ So $\int e^x \sec x+e^x \sec x \tan x d x$ $\frac{d}{d x}(\sec x)=\sec x \tan x$ so this is of the form $\int e^x f(x)+e^x f'(x) d x=e^x f(x)$ $f(x)=\sec x$ $=e^x \sec x+c$