If ∫ (tan x)/(1 + tan x + tan ²x)dx = x - K/(A) tan ⁻¹ ((K tan x + 1)/(A)) + C,(C is a constant of integration) then the ordered pair (K, A) is equal to:

Integral Calculus: JEE Main 2018 Mathematics question

JEE Main 2018 · 16 Apr, Morning shift · Q4007

JEE Main 201816 Apr, Morning shiftMathematicsHardIntegral Calculus369× since 2002

\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}

\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,

is a constant of integration) then the ordered pair (K, A) is equal to :

JEE Main 2018 Integral Calculus Question with Solution (16 Apr Morning shift, Q4007) | JEE Arc

JEE Main 2018 · 16 Apr, Morning shift · Q4007

JEE Main 201816 Apr, Morning shiftMathematicsHardIntegral Calculus369× since 2002

\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}

\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,

is a constant of integration) then the ordered pair (K, A) is equal to :