Solution Let: 2x+1(x−2)(x2+3x+5)=Ax−2+Bx+Cx2+3x+5\frac{2x+1}{(x-2)(x^2+3x+5)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+3x+5}(x−2)(x2+3x+5)2x+1=x−2A+x2+3x+5Bx+C Multiply through: 2x+1=A(x2+3x+5)+(Bx+C)(x−2)2x+1=A(x^2+3x+5)+(Bx+C)(x-2)2x+1=A(x2+3x+5)+(Bx+C)(x−2) Solving gives: A=13,B=−13,C=13A=\frac13,\quad B=-\frac13,\quad C=\frac13A=31,B=−31,C=31 Hence: 2x+1(x−2)(x2+3x+5)=13(x−2)+−x+13(x2+3x+5)\boxed{\frac{2x+1}{(x-2)(x^2+3x+5)}=\frac{1}{3(x-2)}+\frac{-x+1}{3(x^2+3x+5)}}(x−2)(x2+3x+5)2x+1=3(x−2)1+3(x2+3x+5)−x+1