If ${\int }_{ }^{ }\frac{\cos \left(9x\right)+\cos \left(6x\right)}{-1+2\cos \left(5x\right)}dx$$=A\sin \left(4x\right)+B\sin \left(x\right)+C$ where C is constant of integration, then $(A+B)$is equal to

${\int }_{ }^{ }\frac{\cos 9x+\cos 6x}{2\cos 5x-1}dx={\int }_{ }^{ }\frac{2\cos \left(\frac{15x}{2}\right)\cos \left(\frac{3x}{2}\right)}{-1+2(2{\cos }^2\frac{5x}{2}-1)}dx$

$={\int }_{ }^{ }\frac{2\cos \left(\frac{15x}{2}\right).\cos \left(\frac{3x}{2}\right).\cos \left(\frac{5x}{2}\right)}{(4{\cos }^2\frac{5x}{2}-3).\cos \frac{5x}{2}}dx$

$={\int }_{ }^{ }\frac{(2\cos \frac{5x}{2}.\cos \frac{3x}{2}).\cos \left(\frac{15x}{2}\right)}{\cos \left(\frac{15x}{2}\right)}$

$={\int }_{ }^{ }(\cos 4x+\cos x)dx=\frac{1}{4}\sin 4x+\sin x+C$

So, on comparing we get

$A=\frac{1}{4},B=1\Rightarrow (A+B)=\frac{1}{4}+1=\frac{5}{4}$