A) \[{{e}^{-x}}\tan x+c\]
B) \[{{e}^{-x}}\sec x+c\]
C) \[{{e}^{x}}\sec x+c\]
D) \[{{e}^{x}}\tan x+c\]
Correct Answer: C
Solution :
Let \[\int{\frac{1+\tan x}{{{e}^{-x}}\cos x}dx}\] \[=\,\int{{{e}^{x}}\sec x\,dx+\int{{{e}^{x}}\sec x\tan x\,dx}}\] \[=\,{{e}^{x}}\sec x\,-\int{{{e}^{x}}\sec x\,\tan x\,dx}\] \[+\int{{{e}^{x}}\sec x\,\tan x\,dx}+c\] \[=\int{{{e}^{x}}\sec x+c}\]
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