We have,
$$
\begin{aligned}
\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} & =\frac{23}{41} x \\
& +K \log |4 \cos x+5 \sin x|+C
\end{aligned}
$$
On differentiating, we get
$$
\begin{aligned}
& \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x}=\frac{23}{41}+\frac{K(-4 \sin x+5 \cos x)}{4 \cos x+5 \sin x} \\
& 2 \cos x+3 \sin x \\
&= \frac{92 \cos x+115 \sin x-164 K \sin x+205 K \cos x}{41} \\
& \Rightarrow 2 \cos x+3 \sin x=\left(\frac{92+205 K}{41}\right) \cos x \\
&+\left(\frac{115-164 K}{41}\right) \sin x
\end{aligned}
$$
Equating the coefficient of $\cos x$ and $\sin x$, we get
$$
\begin{aligned}
\frac{92+205 K}{41} & =2 \text { and } \frac{115-164 K}{41}=3 \\
\Rightarrow \quad 92+205 K & =82 \\
K & =\frac{-10}{205}=\frac{-2}{41}
\end{aligned}
$$