If $\int \frac{\cos \mathrm{x}+\sin 2\mathrm{x}}{\left(2-{\cos }^2\mathrm{x}\right)(\sin \mathrm{x})}\mathrm{dx}=\int \frac{\mathrm{A}}{\sin \mathrm{x}}\mathrm{dx}+\mathrm{B}\int \frac{\sin \mathrm{x}}{1+{\sin }^2\mathrm{x}}\mathrm{dx}+\mathrm{C}\int \frac{\mathrm{dx}}{1+{\sin }^2\mathrm{x}}$

$\begin{array}{l}\mathrm{I}=\int \frac{\cos \mathrm{x}+\sin 2\mathrm{x}}{\left(2-{\cos }^2\mathrm{x}\right)(\sin \mathrm{x})}\mathrm{dx}\\ \mathrm{I}=\int \frac{(1+2\sin \mathrm{x})\cos \mathrm{x}}{\left(1+{\sin }^2\mathrm{x}\right)(\sin \mathrm{x})}\mathrm{dx}\\ \mathrm{I}=\int \frac{1+2\mathrm{t}}{\left(1+{\mathrm{t}}^2\right)\mathrm{t}}\mathrm{dt},(\text{ Put }\sin \mathrm{x}=\mathrm{t})\end{array}$

Use partial fractions $\frac{1+2\mathrm{t}}{\mathrm{t}\left(1+{\mathrm{t}}^2\right)}=\frac{1}{\mathrm{t}}+\frac{(-\mathrm{t}+2)}{1+{\mathrm{t}}^2}$