Ta có
\(\int \frac{4\sin x+3\cos x}{\sin x+2\cos x} {\rm d}x\)
\(=\int \frac{2\sin x-\cos x}{\sin x+2\cos x} {\rm d}x +2\int {\rm d}x =A+2x+C_{1} ,\, \, C_{1} \in {\rm R}.\)
Đặt \(u=\sin x+2\cos x\Rightarrow {\rm d}u=\left(\cos x-2\sin x\right){\rm d}x\Rightarrow \frac{{\rm d}u}{\cos x-2\sin x} ={\rm d}x.\)
Ta được
\(A=-\int \frac{1}{u} {\rm d}u =\ln \left|u\right|+C_{2} =\ln \left|\sin x+2\cos x\right|+C_{2} ,\, \, C_{2} \in {\rm R}.\)
Vậy \(\int \frac{4\sin x+3\cos x}{\sin x+2\cos x} {\rm d}x =\ln \left|\sin x+2\cos x\right|+2x+C,\, \, C\in {\rm R}\)