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Q.

∫1+tan2x1−tan2xdx=

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a

12ln|1+tan2x1−tan2x|+c

b

12ln|1−tan2x1+tan2x|+c

c

12ln|1+tanx1−tanx|+c

d

12ln|1−tanx1+tanx|+c

answer is C.

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Detailed Solution

∫1+tan2x1−tan2xdx= ∫sec2x1−tan2xdx=∫dt1−t2 where t = tanx =12ln|1+t1−t|+c =12ln|1+tanx1−tanx|+c
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