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Ta có
$\frac{2013\sin x + x\cos x - 4x\sin x - x(4x - 2014)}{x^2 + x\sin x}$
$=\frac{-4x(x+\sin x)+2013(x +\sin x)+x(1+ \cos x)}{x(x + \sin x)}$
$=-4+\frac{2013}{x}+\frac{1+ \cos x}{x + \sin x}$
$=-4+2013 (\ln x)'+(\ln (x+\sin x))'$
Vậy
$\int\frac{2013\sin x + x\cos x - 4x\sin x - x(4x - 2014)}{x^2 + x\sin x}dx$
$=\boxed{-4x +2013 \ln x + \ln (x+\sin x)+ C} $
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