Question

Integrate with respect to x :

$\frac{1}{(x+1)(x+2)}$

Answer 1

$\int \frac{1}{(x+1)(x+2)}dx$

BY partial fraction method;

$\frac{1}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\Rightarrow 1=A(x+2)+B(x+1)\Rightarrow 1=(A+B)x+(2A+B)$

Comparing both side we get,

$A+B=0and2A+B=1\Rightarrow A=1andB=-1\therefore \int \frac{1}{(x+1)(x+2)}dx=\int [\frac{1}{(x+1)}-\frac{1}{(x+2)}]dx=\log (x+1)-\log (x+2)+C=\log \left(\frac{x+1}{x+2}\right)+C$