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∫1+2tanxtanx+secx1/2dx

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a

logsecxsecx−Tanx+C

b

logSecxSecx+Tanx+C

c

logSecxTanx−Secx+C

d

logTanxSecx−Tanx+C

answer is B.

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Detailed Solution

∫1+2tan2x+2tanxsecx1/2dx=∫sec2⁡x+tan2⁡x+2tan⁡csec⁡x1/2dx=∫sec ⁡x+tanxdx=log⁡|sec⁡x+tan⁡x|+log⁡|sec⁡x|+c=log⁡|sec⁡x(sec⁡x+tan⁡x)|+c

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