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a
2tan−1(2tanx)−tan−1(2tanx)+K
b
2tan−1(2tanx)−x+K
c
2tan−1(2tanx)+x2+K
d
2tan−1(2tanx)+x+K
answer is B.
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Detailed Solution
Let J=∫1tan2x+sec2xdx Let J=∫1tan2x+sec2xsec2xsec2xdx=∫sec2xdx2tan2x+11+tan2x Put tanx=t⇒sec2xdx=dt Let J=∫dt2t2+11+t2=∫22t2+1−11+t2dt=22tan−1(2t)−tan−1(t)+KJ=2tan−1(2tanx)−tan−1(tanx)+KJ=2tan−1(2tanx)−x+K