Solution Let: 2x+32(x−2)(x2+2)2=Ax−2+Bx+Cx2+2+Dx+E(x2+2)2\frac{2x+32}{(x-2)(x^2+2)^2}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2}+\frac{Dx+E}{(x^2+2)^2}(x−2)(x2+2)22x+32=x−2A+x2+2Bx+C+(x2+2)2Dx+E Clearing denominators and solving gives: A=1,B=−1,C=−2,D=−6,E=−10A=1,\quad B=-1,\quad C=-2,\quad D=-6,\quad E=-10A=1,B=−1,C=−2,D=−6,E=−10 Hence: 2x+32(x−2)(x2+2)2=1x−2−x+2x2+2−6x+10(x2+2)2\boxed{\frac{2x+32}{(x-2)(x^2+2)^2}=\frac{1}{x-2}-\frac{x+2}{x^2+2}-\frac{6x+10}{(x^2+2)^2}}(x−2)(x2+2)22x+32=x−21−x2+2x+2−(x2+2)26x+10