\[ \int \frac{2 - \tan x}{3 + \tan x} \, dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx \]
Let:
\[ 2 \cos x - \sin x = A(3 \cos x + \sin x) + B(\cos x - 3 \sin x) \]
Equating coefficients:
\[ 3A + B = 2, \quad A - 3B = -1 \]
Solving these equations:
\[ A = \frac{1}{2}, \quad B = \frac{1}{2} \]
Thus:
\[ \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx = \frac{x}{2} + \frac{1}{2} \ln |3 \cos x + \sin x| + C \]
Simplify:
\[ = \frac{1}{2} \left( x + \ln |3 \cos x + \sin x| \right) + C \]
Rewriting in the given form:
\[ = \frac{1}{2} \left( \alpha x + \ln |\beta \sin x + \gamma \cos x| \right) + C \]
From the equation:
\[ \alpha = 1, \quad \beta = 1, \quad \gamma = 3 \]
Finally:
\[ \alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4 \]
Ans: Option (3): 4.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32