If $\int \frac{5\cot x+1}{(\cot x-1)(\cot x-2){\sin }^{2}x}dx=6\log \left|f\left(x\right)\right|+11\log \left|g\left(x\right)\right|+c, then \left(f\right(x), g(x\left)\right)=$

$$\begin{gathered} \int \frac{5 \cot \mathrm{x}+1}{(\cot \mathrm{x}-1)(\cot \mathrm{x}-2){\sin }^2\mathrm{x}}\mathrm{dx} \\ =\int \frac{5\mathrm{t}+1}{(\mathrm{t}-1)(\mathrm{t}-2)}(-\mathrm{dt}) \\ \mathrm{Put} \cot \mathrm{x}=\mathrm{t}\Rightarrow -{\mathrm{cossec}}^2\mathrm{x} \mathrm{dx}=\mathrm{dt} \\ =\int \left[\frac{-6}{\mathrm{t}-1}+\frac{11}{\mathrm{t}-2}\right]\mathrm{dt} \\ =-6 \log \left|\mathrm{t}-1\right|+11 \log \left|\mathrm{t}-2\right| \\ = 6 \log \frac{1}{\mathrm{t}-1}+11 \log (\mathrm{t}-2) \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\cot \mathrm{x}-1} \\ \mathrm{g}\left(\mathrm{x}\right)= \cot \mathrm{x}-2 \end{gathered}$$