Ta có
$I = \displaystyle \int \dfrac{4\sin x + 3 \cos x}{\sin x + 2 \cos x} dx$
$= \displaystyle \int \dfrac{2(\sin x + 2 \cos x) - (\cos x - 2\sin x)}{\sin x + 2 \cos x}dx$
$= \displaystyle \int \left( 2 - \dfrac{\cos x - 2\sin x}{\sin x + 2\cos x}\right) dx$
$= 2x - \displaystyle \int \dfrac{(\sin x + 2 \cos x)'}{\sin x + 2 \cos x}dx$
$= 2x - \displaystyle \int \dfrac{d(\sin x + 2 \cos x)}{\sin x + 2 \cos x}$
$= 2 x - \ln|\sin x + 2\cos x| + c$.
