∫(sec⁡x+tan⁡x)2dx=

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2(sec⁡x+tan⁡x)−x+c

1/3(sec⁡x+tan⁡x)3+c

sec⁡x(sec⁡x+tan⁡x)+c

2(sec⁡x+tan⁡x)+c

answer is A.

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Detailed Solution

∫(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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