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$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{1+\sin x}{1+3\cos x}dx=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{1}{1+3\cos x}dx- \dfrac{1}{3}\int\limits_{0}^{\frac{\pi}{2}}\dfrac{-3\sin x}{1+3\cos x}dx$
$I=I_1- \dfrac{1}{3}\ln |1+3\cos x|_{0}^{\pi/2}$
$I=I_1+\dfrac{2\ln 2}{3}$
Đặt $t = \tan \dfrac{x}{2}\Rightarrow dt = \dfrac{1}{2}\dfrac{1}{\cos^2 \dfrac{x}{2}}dx= \dfrac{1}{2}(1+t^2)dx$
Suy ra
$I_1 =\displaystyle \int\limits_{0}^{1}\dfrac{1}{1+3\dfrac{1-t^2}{1+t^2}}.\dfrac{2}{1+t^2}dt$
$I_1 =\displaystyle \int\limits_{0}^{1}\dfrac{1}{2-t^2}dt=\dfrac{1}{2\sqrt 2}\ln \left| {\dfrac{\sqrt 2 +t}{\sqrt 2 -t}} \right|_{0}^{1}=\dfrac{1}{2\sqrt 2}\ln \left ( {\dfrac{\sqrt 2 +1}{\sqrt 2 -1}} \right )$
Vậy $\boxed{I=\dfrac{1}{2\sqrt 2}\ln \left ( {\dfrac{\sqrt 2 +1}{\sqrt 2 -1}} \right )+\dfrac{2\ln 2}{3}.}$
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