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B
exsecx+C
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C
exsinx+C
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D
extanx+C
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Solution
The correct option is Bexsecx+C Let I=∫exsecx(1+tanx)dx=∫ex(secx+secxtanx)dx Also, let secx=f(x)⇒secxtanx=f′(x) We know that, ∫ex{f(x)+f′(x)}=exf(x)+C ∴I=exsecx+C