∫1/((x+1)(x+2))dx=

The correct answer is Option (2) → $\log\frac{x+1}{x+2}+C;$ Where C is arbitrary constant of integration

$∫\frac{1}{(x+1)(x+2)}dx=∫\frac{(x+2)-(x+1)}{(x+1)(x+2)}dx$

$=∫\frac{1}{(x+1)}-\frac{1}{(x+2)}dx$

$=\log(x+1)-\log(x+2)+C$

$=\log\left|\frac{(x+1)}{(x+2)}\right|+C$