Is everything clear about integrals of sines and cosines?

Dividing the numerator and the denominator of $\frac{\sin^2x\cos^2 x}{(\cos^3x+\sin^3x)^2}$ by $\cos^6x,$ we obtain the expression $\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}.$

Then $I=\int\frac{\sin^2x\cos^2 x}{(\sin ^3x+\cos^3x)^2}dx=\int\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}dx=\frac{1}{3}\int\frac{d\tan^3x}{(1+\tan^3x)^2}=-\frac{1}{3(1+\tan^3x)}+C.$ Multiplying the last fraction by $\frac{\cos^3x}{\cos^3x},$ we get that $I=-\frac{1}{3}\frac{\cos^3x}{\cos^3x+\sin^3x}+C.$ Factoring the denominator and using the identity $\sin^2 x+\cos^2x=1$ we get that $I=-\frac{1}{3}\frac{\cos^3x}{(\cos x+\sin x)(\cos^2x-\sin x\cos x+\sin^2 x)}+C=-\frac{1}{3}\frac{\cos^3x}{(\cos x+\sin x)(1-\sin x\cos x)}+C.$ Therefore, $A=B=1.$ So the answer is $\boxed{A+B=2}.$