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$I = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x + (x + \sin x)\sin x}{\sin^2 x (1 + \sin x)} dx$
$I = \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x}{\sin^2 x}dx+ \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{1}{1 + \sin x} dx$
$I=I_1+I_2$
Trong đó
$I_1=\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{xdx}{\sin^2x}=\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left
(\cot x -\cot x+ \frac{x}{\sin^2x} \right )dx$
$=
\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \cot xdx-
\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left (\cot x-
\frac{x}{\sin^2x} \right )dx$
$=
\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{d(\sin x)}{\sin x}-
\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left (x\cot x \right )'dx$
$=\left[ {\ln |\sin x|-x\cot x} \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} $
$=\displaystyle{\frac{1}{\sqrt 3}\pi}$
$I_2= \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{1}{1 + \sin x} dx$
$=2 \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{(\sin\frac{x}{2}+\cos \frac{x}{2})(\sin\frac{x}{2})'-(\sin\frac{x}{2}+\cos \frac{x}{2})'(\sin\frac{x}{2})}{(\sin\frac{x}{2}+\cos \frac{x}{2})^2} dx$
$=2 \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \left ( \frac{\sin\frac{x}{2}}{\sin\frac{x}{2}+\cos \frac{x}{2}} \right )'
dx$
$=\left[ { 2\frac{\sin\frac{x}{2}}{\sin\frac{x}{2}+\cos \frac{x}{2}} } \right]_{\frac{\pi}{3}}^{\frac{2\pi}{3}} $
$4-2\sqrt 3$
Vậy $\boxed{I=\frac{1}{\sqrt 3}\pi+4-2\sqrt 3}$
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