Decompose rational expressions

Write $\dfrac4{(x-2)(x+2)}=\dfrac A{x-2}+\dfrac B{x+2}$. Clearing denominators gives $4=A(x+2)+B(x-2)$. Substituting $x=2$ gives $A=1$ and $x=-2$ gives $B=-1$. Thus $\dfrac4{x^2-4}=\dfrac1{x-2}-\dfrac1{x+2}$, for $x\ne\pm2$.

Write $\dfrac{x^2-11}{(x+2)^2(3x-1)}=\dfrac A{x+2}+\dfrac B{(x+2)^2}+\dfrac C{3x-1}$. Then $x^2-11=A(x+2)(3x-1)+B(3x-1)+C(x+2)^2$. At $x=1/3$, $-98/9=(49/9)C$, so $C=-2$. At $x=-2$, $-7=-7B$, so $B=1$. At $x=0$, $-11=-2A-1-8$, so $A=1$. Therefore the answer is $\dfrac1{x+2}+\dfrac1{(x+2)^2}-\dfrac2{3x-1}$, for $x\ne-2,1/3$.

The fraction is improper. Polynomial division subtracts $x(x^3+2x)=x^4+2x^2$, leaving $1-2x^2$. Thus $\dfrac{x^4+1}{x^3+2x}=x+\dfrac{1-2x^2}{x(x^2+2)}$.

Write $\dfrac{1-2x^2}{x(x^2+2)}=\dfrac A x+\dfrac{Bx+C}{x^2+2}$. Then $1-2x^2=A(x^2+2)+(Bx+C)x$. Comparing coefficients gives $C=0$, $2A=1$, and $A+B=-2$, hence $A=1/2$ and $B=-5/2$. The original fraction is $x+\dfrac1{2x}-\dfrac{5x}{2(x^2+2)}$, for real $x\ne0$.