$∫e^x(tan\, x + log_e secx)dx=$
Answer & explanation
Correct answer: option 1
The correct answer is option (1) → $e^x\log_e\sec x+C$
I=$∫e^x(\tan x + \log_e \sec x)dx$
$f'(x)=\tan x,f(x)=\log_e \sec x$
so $I = e^xf(x)+C$
$=e^x\log_e\sec x+C$
$∫e^x(tan\, x + log_e secx)dx=$
Correct answer: option 1
The correct answer is option (1) → $e^x\log_e\sec x+C$
I=$∫e^x(\tan x + \log_e \sec x)dx$
$f'(x)=\tan x,f(x)=\log_e \sec x$
so $I = e^xf(x)+C$
$=e^x\log_e\sec x+C$