$$
\begin{aligned}
\text { } & \int \frac{d x}{3 \cos x-4 \sin x+5} \\
= & \int \frac{d x}{\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)-4\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+5} \\
= & \int \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{3\left(1-\tan ^2 \frac{x}{2}\right)-4\left(2 \tan \frac{x}{2}\right)+5\left(1+\tan ^2 \frac{x}{2}\right)} \\
= & \int \frac{\sec ^2 \frac{x}{2} d x}{2 \tan ^2 \frac{x}{2}-8 \tan \frac{x}{2}+8}=\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2} d x}{\tan ^2 \frac{x}{2}-4 \tan \frac{x}{2}+4} \\
= & \frac{1}{2} \int \frac{\sec ^2 \frac{x}{2} d x}{\left(\tan ^{\frac{x}{2}}-2\right)^2}
\end{aligned}
$$

Let $\tan \frac{x}{2}-2=u, \frac{1}{2} \sec ^2 \frac{x}{2} d x=d u$
$$
\begin{aligned}
& =\int \frac{d u}{u^2}=\frac{-1}{u}+C \\
& =\frac{-1}{\tan \frac{x}{2}-2}+C=\frac{1}{2-\tan \frac{x}{2}}+C .
\end{aligned}
$$