The correct option is C ln |sec x(sec x+tan x)|+c
∫{1+2 tan x(tan x+sec x)}12 dx
∫{1+2 tan2 x+2 tan x sec x}12 dx
usng trigonometric identity sec2 x−tan2 x=1
∫{sec2 x+tan2 x+2 tan x sec x}12 dx
=∫(sec x+tan x) dx
=ln sec x+ln (sec x+tan x)+c
=ln |sec x(sec x+tan x)|+c