Precalculus, Chapter 7, 7.4, Section 7.4, Problem 36

x^2/(x^4-2x^2-8)
Let's factorize the denominator,
x^4-2x^2-8=x^4-4x^2+2x^2-8
=x^2(x^2-4)+2(x^2-4)
=(x^2+2)(x^2-4)
=(x^2+2)(x+2)(x-2)
:.x^2/(x^4-2x^2-8)=x^2/((x+2)(x-2)(x^2+2))
Let x^2/((x+2)(x-2)(x^2+2))=A/(x+2)+B/(x-2)+(Cx+D)/(x^2+2)
=(A(x-2)(x^2+2)+B(x+2)(x^2+2)+(Cx+D)(x+2)(x-2))/((x+2)(x-2)(x^2+2))
=(A(x^3+2x-2x^2-4)+B(x^3+2x+2x^2+4)+(Cx+D)(x^2-4))/((x+2)(x-2)(x^2+2))
=(A(x^3-2x^2+2x-4)+B(x^3+2x^2+2x+4)+Cx^3-4Cx+Dx^2-4D)/((x+2)(x-2)(x^2+2))
=(x^3(A+B+C)+x^2(-2A+2B+D)+x(2A+2B-4C)-4A+4B-4D)/((x+2)(x-2)(x^2+2))
Now,
x^2=x^3(A+B+C)+x^2(-2A+2B+D)+x(2A+2B-4C)-4A+4B-4D
Equating the coefficients of like terms,
A+B+C=0 --- equation 1
-2A+2B+D=1 ------ equation 2
2A+2B-4C=0 ------ equation 3
-4A+4B-4D=0 ----- equation 4
Now we have to solve the above four equations to find the solutions of A,B,C and D.
Rewrite equation 1 ,
A+B=-C
Substitute above in equation 3,
2(A+B)-4C=0
2(-C)-4C=0
-2C-4C=0
-6C=0
C=0
Rewrite equation 2 as,
-2A+2B=1-D
Substitute the above in equation 4,
-4A+4B-4D=0
2(-2A+2B)-4D=0
2(1-D)-4D=0
2-2D-4D=0
2-6D=0
D=2/6
D=1/3
Plug the values of C in equation 3 ,
2A+2B-4(0)=0
2A+2B=0
2(A+B)=0
A+B=0 ------ equation 5
Plug the value of D in equation 4,
-4A+4B-4(1/3)=0
-4A+4B=4/3
4(-A+B)=4/3
-A+B=1/3 ---- equation 6
Solve equations 5 and 6 to find the solutions of A and B,
Add the equations 5 and 6.
2B=1/3
B=1/6
Plug the value of B in equation 5,
A+1/6=0
A=-1/6
:.x^2/(x^4-2x^2-8)=(-1/6)/(x+2)+(1/6)/(x-2)+(0*x+1/3)/(x^2+2)
x^2/(x^4-2x^2-8)=-1/(6(x+2))+1/(6(x-2))+1/(3(x^2+2))