QJEEM24Ma8f0f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma8f0f Solution: Let , where is the constant of integration. Then is equal to :

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

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

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

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