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(secx+tanx)2dx=

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a
2(sec⁡x+tan⁡x)−x+c
b
1/3(sec⁡x+tan⁡x)3+c
c
sec⁡x(sec⁡x+tan⁡x)+c
d
2(sec⁡x+tan⁡x)+c

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detailed solution

Correct option is A

∫(sec⁡x+tan⁡x)2dx=∫sec2⁡x+tan2⁡x+2sec⁡xtan⁡xdx=∫2sec2⁡x−1+2sec⁡xtan⁡xdx=2(sec⁡x+tan⁡x)−x+c


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