\int\frac{1+\tan^2 x}{1-\tan^2 x} dx=

Correct Answer is : 12 log(1+tanx1−tanx) + c

\log (\frac{1-\tan x}{1+\tan x}) + c

\log (\frac{1+\tan x}{1-\tan x}) + c

{1/2} \log (\frac{1-\tan x}{1+\tan x}) + c

1/2 \log (\frac{1+\tan x}{1-\tan x}) + c

AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Section: Mathematics

Question : 70 of 160

Marks: +1, -0

∫

1+tan2x

1−tan2x

dx=

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