Solution First factor: (x+1)(x2+5x+6)=(x+1)(x+2)(x+3)(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)(x+1)(x2+5x+6)=(x+1)(x+2)(x+3) Let: x2+4x+5(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3\frac{x^2+4x+5}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}(x+1)(x+2)(x+3)x2+4x+5=x+1A+x+2B+x+3C Solving gives A=1A=1A=1, B=−1B=-1B=−1, C=1C=1C=1. x2+4x+5(x+1)(x2+5x+6)=1x+1−1x+2+1x+3\boxed{\frac{x^2+4x+5}{(x+1)(x^2+5x+6)}=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+3}}(x+1)(x2+5x+6)x2+4x+5=x+11−x+21+x+31