Given that ∫ (3 x+4 x)/(2 x+5 x) d x=(α x)/(29)+(β)/(29) |2 x+5...

Let, $$\begin{aligned} & \mathrm{I}=\int \frac{3 \cos x+4 \sin x}{2 \cos x+5 \sin x} d x \\ & 3 \cos x+4 \sin x=\mathrm{A}(2 \cos x+5 \sin x) \\ & +\mathrm{B}\left[\frac{d}{d x}(2 \cos x+5 \sin x)\right] \\ & 3 \cos x+4 \sin x=\mathrm{A}(2 \cos x+5 \sin x) \\ & +\mathrm{B}(-2 \sin x+5 \cos x) \end{aligned}$$ Comparing co-efficient of $\sin x and \cos x$$\begin{aligned} & 3=2 A+5 B \qquad ....(i)\\ & 4=5 A-2 B \qquad ....(ii) \end{aligned}$$$Solving (i) and (ii) then we get,$$\begin{aligned} & \mathrm{A}=\frac{26}{29}, \mathrm{ B}=\frac{7}{29} \\ & \mathrm{I}=\frac{\int \frac{26}{29}(2 \cos x+5 \sin x)+\frac{7}{29}(-2 \sin x+5 \cos x)}{2 \cos x+5 \sin x} d x \\ &=\int \frac{26}{29} d x+\frac{7}{29} \int \frac{(-2 \sin x+5 \cos x)}{2 \cos x+5 \sin x} d x \\ &=\frac{26}{29} x+\frac{7}{29} \text{ In }(2 \cos x+5 \sin x)+c \\ & \alpha=26, \beta=7[\text{ [on comparing] } \end{aligned}$$\alpha=26$