\[\begin{array}{l}
I = \int {\dfrac{{x\sin x}}{{1 + 2\cos 2x}}dx} = \dfrac{1}{2}\int {\dfrac{{x\sin x}}{{\dfrac{1}{2} + \cos 2x}}dx} \\
\dfrac{1}{2} + \cos 2x = \cos 2x + \cos \dfrac{\pi }{3} = 2\cos \left( {x + \dfrac{\pi }{6}} \right)\cos \left( {x - \dfrac{\pi }{6}} \right)\\
= 2\left[ {\cos x\cos \dfrac{\pi }{6} - \sin x\sin \dfrac{\pi }{6}} \right]\left[ {\cos x\cos \dfrac{\pi }{6} + \sin x\sin \dfrac{\pi }{6}} \right]\\
= 2\left( {{{\cos }^2}x{{\cos }^2}\dfrac{\pi }{6} - {{\sin }^2}x{{\sin }^2}\dfrac{\pi }{6}} \right)\\
= \dfrac{3}{2}{\cos ^2}x - \dfrac{1}{2}{\sin ^2}x = \dfrac{3}{2}(1 - {\sin ^2}x) - \dfrac{1}{2}{\sin ^2}x = \dfrac{3}{2} - 2{\sin ^2}x\\
I = \int {\dfrac{{x\sin x}}{{3 - 4{{\sin }^2}x}}dx} = \int {\dfrac{{x{{\sin }^2}x}}{{3\sin x - 4{{\sin }^3}x}}dx} = \int {\dfrac{{x{{\sin }^2}x}}{{\sin 3x}}dx}
\end{array}\]
\[I = \dfrac{1}{2}\int {\dfrac{{x(1 - \cos 2x)}}{{\sin 3x}}dx} \Rightarrow \boxed{2I = \int {\dfrac{x}{{\sin 3x}}dx} - \int {\dfrac{{\cos 2x}}{{\sin 3x}}dx} }\]
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