Chapter 5, Systems of Equations and Inequalities - Chapter 5 Review - Exercises - Page 479: 35

$\frac{3x^2-x+6}{(x^2+2)^2}=\frac{3}{x^2+2}+\frac{-x}{(x^2+2)^2}$,

$\frac{3x^2-x+6}{(x^2+2)^2}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{(x^2+2)^2}$, $(Ax+B)(x^2+2)+Cx+D$, $Ax^3+Bx^2+(2A+C)x+(2B+D)=3x^2-x+6$, thus, $A=0, B=3, C=-1, D=0$, Therefore, $\frac{3x^2-x+6}{(x^2+2)^2}=\frac{3}{x^2+2}+\frac{-x}{(x^2+2)^2}$,