Precalculus, Chapter 7, 7.4, Section 7.4, Problem 35

(2x^2+x+8)/(x^2+4)^2
Let(2x^2+x+8)/(x^2+4)^2=(Ax+B)/(x^2+4)+(Cx+D)/(x^2+4)^2
(2x^2+x+8)/(x^2+4)^2=((Ax+B)(x^2+4)+Cx+D)/(x^2+4)^2
(2x^2+x+8)/(x^2+4)^2=(Ax^3+4Ax+Bx^2+4B+Cx+D)/(x^2+4)^2
:.(2x^2+x+8)=Ax^3+4Ax+Bx^2+4B+Cx+D
2x^2+x+8=Ax^3+Bx^2+(4A+C)x+4B+D
Equating the coefficients the like terms,
A=0
B=2
4A+C=1
4B+D=8
Plug the value of the A and B in the above equations,
4(0)+C=1
C=1
4(2)+D=8
8+D=8
D=8-8
D=0
:.(2x^2+x+8)/(x^2+4)^2=2/(x^2+4)+x/(x^2+4)^2
Now let's check it algebraically,
RHS=2/(x^2+4)+x/(x^2+4)^2
=(2(x^2+4)+x)/(x^2+4)^2

=(2x^2+8+x)/(x^2+4)^2

=(2x^2+x+8)/(x^2+4)^2
=LHS
Hence it is verified.