\int \frac{1}{1+3 \sin ^2 x+8 \cos ^2 x} \dx=

Correct Answer is : 16tan−1(2tanx3)+C

\frac{1}{6} \tan ^{-1} (\frac{2 \tan x}{3}) +C

6 \tan ^{-1} (\frac{2 \tan x}{3}) + {\C}

\frac{1}{6} \tan ^{-1}(2 \tan x)+C

\tan ^{-1} (\frac{2 \tan x}{3}) +C

KCET 2023 Math Solved Paper

Question : 23 of 60

Marks: +1, -0

∫‌

1+3sin‌2x+8cos2x

‌dx=

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