$$I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$$

Put, $$2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)$$

$$\begin{aligned} & a+3 b=2 \quad \text{.... (i)}\\ & -3 a+b=-1 \quad \text{.... (ii)} \end{aligned}$$

From equation (i) and (ii) $$a=b=\frac{1}{2}$$

$$I=\frac{1}{2} \int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\frac{1}{2} \int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x$$

$$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x= \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$$

On comparing, we get

$$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x=\frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$$

On comparing, we get

$$\begin{aligned} & \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \\ & \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) \end{aligned}$$

$$\begin{aligned} & \alpha=1, \beta=1, \gamma=3 \\ & \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4 \end{aligned}$$