The correct option is C logx+1x+2+C
Let
1(x+1)(x+2)=Ax+1+Bx+2
1=A(x+2)+B(x+1)⇒Ax+Bx+2A+B=1
Comparing the coefficient of x and constant on both sides,
A+B=0 and 2A+B=1
Solving the above two equations, we get
A=1 and B=−1
⇒∫dx(x+1)(x+2)=∫(1(x+1)−1(x+2))dx
=log(x+1)–log(x+2)+C=logx+1x+2+C