Solution Let: 2x2+3x+3(x+1)(x2+1)=Ax+1+Bx+Cx2+1\frac{2x^2+3x+3}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}(x+1)(x2+1)2x2+3x+3=x+1A+x2+1Bx+C Multiply by (x+1)(x2+1)(x+1)(x^2+1)(x+1)(x2+1): 2x2+3x+3=A(x2+1)+(Bx+C)(x+1)2x^2+3x+3=A(x^2+1)+(Bx+C)(x+1)2x2+3x+3=A(x2+1)+(Bx+C)(x+1) Solve to get: A=1,B=1,C=2A=1,\quad B=1,\quad C=2A=1,B=1,C=2 Hence: 2x2+3x+3(x+1)(x2+1)=1x+1+x+2x2+1\boxed{\frac{2x^2+3x+3}{(x+1)(x^2+1)}=\frac{1}{x+1}+\frac{x+2}{x^2+1}}(x+1)(x2+1)2x2+3x+3=x+11+x2+1x+2