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Let $\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _{\mathrm{e}}|\beta \sin x+\gamma \cos x|\right)+C$, where $C$ is the constant of integration. Then $\alpha+\frac{\gamma}{\beta}$ is equal to :
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4
$\begin{aligned} & \int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \\ & 2 \cos x-\sin x=A(3 \cos x+\sin x)+B(\cos x-3 \sin x) \\ & 3 A+B=2 \\ & A-3 B=-1\end{aligned}$
$\begin{aligned} & \Rightarrow A=\frac{1}{2}, B=\frac{1}{2} \\ & \therefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \\ & =\frac{x}{2}+\frac{1}{2} \ln |3 \cos x+\sin x|+C \\ & =\frac{1}{2}(x+\ln |3 \cos x+\sin x|)+C \\ & =\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C \\ & \alpha=1, \beta=1, \gamma=3 \\ & \therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4\end{aligned}$
$\begin{aligned} & \Rightarrow A=\frac{1}{2}, B=\frac{1}{2} \\ & \therefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x \\ & =\frac{x}{2}+\frac{1}{2} \ln |3 \cos x+\sin x|+C \\ & =\frac{1}{2}(x+\ln |3 \cos x+\sin x|)+C \\ & =\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C \\ & \alpha=1, \beta=1, \gamma=3 \\ & \therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4\end{aligned}$
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