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$\int {1 + 2 \tan x (\tan x + \sec x)}^{1 / 2} d x =$

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log⁡(sec⁡x+tan⁡x)+c

log⁡(sec⁡x+tan⁡x)1/2+c

log⁡sec⁡x(sec⁡x+tan⁡x)+c

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detailed solution

Correct option is C

∫1+2tan2⁡x+2tan⁡xsec⁡x1/2dx=∫sec2⁡x+tan2⁡x+2tan⁡xsec⁡x1/2dx=∫(sec⁡x+tan⁡x)dx=log⁡(sec⁡x+tan⁡x)+log⁡sec⁡x+c=log⁡sec⁡x(sec⁡x+tan⁡x)+c

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∫{1+2tan⁡x(tan⁡x+sec⁡x)}1/2dx=

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$\int {1 + 2 \tan x (\tan x + \sec x)}^{1 / 2} d x =$