Let int frac{2-tan x}{3+tan x} dx = frac{1}{2 | vedclass

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(C) We have $I = \int \frac{2-	an x}{3+	an x} dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} dx$.Let $2 \cos x - \sin x = A(3 \cos x + \sin x

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$4$

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(C) We have $I = \int \frac{2-	an x}{3+	an x} dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} dx$.Let $2 \cos x - \sin x = A(3 \cos x + \sin x) + B(-3 \sin x + \cos x)$.Equating the coefficients of $\cos x$ and $\sin x$:$3A + B = 2$ and $A - 3B = -1$.Solving these equations,we get $A = \frac{1}{2}$ and $B = \frac{1}{2}$.Thus,$I = \int \left( \frac{1}{2} + \frac{1}{2} \frac{-3 \sin x + \cos x}{3 \cos x + \sin x} ight) dx$.$I = \frac{1}{2} x + \frac{1}{2} \ln |3 \cos x + \sin x| + C$.Comparing this with $\frac{1}{2}(\alpha x + \ln |\beta \sin x + \gamma \cos x|) + C$,we get $\alpha = 1$,$\beta = 1$,and $\gamma = 3$.Therefore,$\alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4$.

Let $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$,where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to:

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