$∫\frac{3+2\cos x}{(2+3\cos x)^2}dx$ is equal to

$(\frac{\sin x}{3\cos x+2})+c$

Let $I=∫\frac{3+2\cos x}{(2+3\cos x)^2}dx$

Multiplying numerator and denominator by cosec²x, we get 

$I=∫\frac{3cosec^2x+2\cot x\,cosecx}{(2cosecx+3\cot x)^2}dx=-∫\frac{-3cosec^2x-2cotx\,cosecx}{(2cosecx+3\cot x)^2}$

$=\frac{1}{2cosecx+3cotx}+c=(\frac{sinx}{2+3cosx})+c$

Hence (A) is the correct answer.

Alternative:

$I=∫\frac{3sin^2x+3cos^2x+2cosx}{(2+3cosx)^2}dx$

$=∫\frac{cosx}{(2+3cosx)}dx+∫\frac{3sinx.sinx}{(2+3cosx)^2}dx$

$=∫\frac{cosx}{2+3cosx}dx+\frac{sinx}{2+3cosx}-\int\frac{cosx}{2+3cosx}dx=\frac{sinx}{2+3cosx}+c$