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$∫e^x(tan\, x + log_e secx)dx=$ |
$e^xlog_esecx+C$ $log_e\, sec\, x+C$ $e^xtan\, x +C$ $e^xsec\, x +C$ |
$e^xlog_esecx+C$ |
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 $I = e^xf(x)+C$ $=e^x\log_e\sec x+C$ |
