If $∫\frac{dx}{(x+1)(x+2)}=∫\frac{A}{x+1}dx+∫\frac{B}{x+2}dx,$ then $A+B =$

2 1 0 -1

0

The correct answer is Option (3) → 0 $∫\frac{dx}{(x+1)(x+2)}=∫\frac{(x+2)-(x+1)}{(x+1)(x+2)}dx$ $=∫\frac{1}{(x+1)}-\frac{1}{(x+2)}dx$ $⇒A=1,B=-1$ so $A+B=0$