Solution Let: 2x+3(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3\frac{2x+3}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}(x+1)(x+2)(x+3)2x+3=x+1A+x+2B+x+3C Then: 2x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)2x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)2x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) Solving gives A=12A=\tfrac12A=21, B=1B=1B=1, C=−32C=-\tfrac32C=−23. 2x+3(x+1)(x+2)(x+3)=12(x+1)+1x+2−32(x+3)\boxed{\frac{2x+3}{(x+1)(x+2)(x+3)}=\frac{1}{2(x+1)}+\frac{1}{x+2}-\frac{3}{2(x+3)}}(x+1)(x+2)(x+3)2x+3=2(x+1)1+x+21−2(x+3)3