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Ta có
$ \dfrac{x +\sin ^{2}x}{1 +\cos2x}= \dfrac{1}{2}.\dfrac{x +1-\cos^2 x}{\cos^2 x}= \dfrac{1}{2}\left[ {-1+\tan x +\dfrac{x }{\cos^2 x} +\dfrac{1 }{\cos^2 x}-\tan x} \right]$
$= \dfrac{1}{2}\left[ {(-x)'+(x \tan x)' +(\tan x)'- (\ln |\cos x|)'} \right]$
Vậy
$I= \int\limits_{0}^{\frac{\pi}{3}} \dfrac{x +\sin ^{2}x}{1 +\cos2x}dx = \dfrac{1}{2}\left[ {-x+x \tan x +\tan x- \ln |\cos x|} \right]_{0}^{\frac{\pi}{3}}=\boxed{\dfrac{3\sqrt 3+(\sqrt 3 -1)\pi -3\ln 2}{6}}$
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