MathematicsHard64×since 2004Q4007If ∫tanx1+tanx+tan2xdx=x−KAtan−1\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}∫1+tanx+tan2xtanxdx=x−AKtan−1 (K tanx+1A)+C,(C \left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,(AKtanx+1)+C,(C is a constant of integration) then the ordered pair (K, A) is equal to :A(2, 1)B(−-−2, 3)C(2, 3)D(−-−2, 1)Check answerSkip