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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

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detailed solution

Correct option is B

∫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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