Mathematics · 2018
JEE Main · 16 April 2018, Shift 1 · Q74
If ∫ (tan x)/(1+ tan x+ tan^2 x) d x=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…
If $\displaystyle \int \frac{\tan x}{1+\tan x+\tan ^2 x} \mathrm{~d} x=x-$ $\displaystyle \frac{\mathrm{K}}{\sqrt{\mathrm{A}}} \tan ^{-1}\left(\frac{\mathrm{~K} \tan x+1}{\sqrt{\mathrm{~A}}}\right)+\mathrm{C}$, (C is a constant of integration), then the ordered pair (K, A) is equal to :
Official answer
(3)
$\displaystyle (2,3)$
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.