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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 $A+B=0$ |
