$\int \frac{9\cos \mathrm{x}-\sin \mathrm{x}}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}$
Let $\mathrm{Nr}\equiv \mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{Dr}\right)+\mathrm{B}\left(\mathrm{Dr}\right)$
Take $9\cos \mathrm{x}-\sin \mathrm{x}$
$\begin{array}{l}\equiv \mathrm{A}\frac{\mathrm{d}}{\mathrm{dx}}(5\cos \mathrm{x}+4\sin \mathrm{x})+\mathrm{B}(5\cos \mathrm{x}+4\sin \mathrm{x})-\left(1\right)\\ 9\cos \mathrm{x}-\sin \mathrm{x}\\ \equiv \mathrm{A}(-5\sin \mathrm{x}+4\cos \mathrm{x})+\mathrm{B}(5\cos \mathrm{x}+4\sin \mathrm{x})\\ \end{array}$
Comparing the coefficients of sinx terms and cosx terms on both sides we get
$\begin{array}{l}-5\mathrm{A}+4\mathrm{B}=-1 ...\left(1\right)\\ 4\mathrm{A}+5\mathrm{B}=9 ...\left(2\right)\end{array}$
Soving (1) & (2) we get
$$\begin{gathered} \mathrm{A} \mathrm{B} 1 \\ 4 1 -5 4 \\ 5 -9 4 5 \\ \Rightarrow \frac{\mathrm{A}}{ -36-5}=\frac{\mathrm{B}}{4-45}=\frac{1}{-25-16} \\ \Rightarrow \frac{\mathrm{A}}{-41} =\frac{\mathrm{B}}{-41}=\frac{1}{-41} \end{gathered}$$
A = 1      B = 1 in equation (1)
$\begin{array}{l}9\cos \mathrm{x}-\sin \mathrm{x}=\mathrm{1. }\frac{\mathrm{d}}{\mathrm{dx}}(5\cos \mathrm{x}+4\sin \mathrm{x})+1(5\cos \mathrm{x}+4\sin \mathrm{x})\\ \\ \int \frac{9\cos \mathrm{x}-\sin \mathrm{x}}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}\\ =\int \frac{\frac{\mathrm{d}}{\mathrm{dx}}(5\cos \mathrm{x}+4\sin \mathrm{x})+(5\cos \mathrm{x}+4\sin \mathrm{x})}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}\\ \therefore =\int \frac{\frac{\mathrm{d}}{\mathrm{dx}}(5\cos \mathrm{x}+4\sin \mathrm{x})}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}+\int \frac{5\cos \mathrm{x}+4\sin \mathrm{x}}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}\\ =\log |5\cos \mathrm{x}+4\sin \mathrm{x}|+\mathrm{x}+\mathrm{c}\\ \therefore \int \frac{9\cos \mathrm{x}-\sin \mathrm{x}}{5\cos \mathrm{x}+4\sin \mathrm{x}}\mathrm{dx}\\ =\log (5\cos \mathrm{x}+4\sin \mathrm{x})+\mathrm{x}+\mathrm{c}\\ \end{array}$