Precalculus, Chapter 7, 7.4, Section 7.4, Problem 37

x/(16x^4-1)
Let's factorize the denominator,
16x^4-1=(4x^2)^2-1
=(4x^2+1)(4x^2-1)
=(4x^2+1)(2x+1)(2x-1)
Let x/(16x^4-1)=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)
x/(16x^4-1)=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+1))
x/(16x^4-1)=(A(8x^3+2x-4x^2-1)+B(8x^3+2x+4x^2+1)+(Cx+D)(4x^2-1))/((2x+1)(2x-1)(4x^2+1))
x/(16x^4-1)=(A(8x^3-4x^2+2x-1)+B(8x^3+4x^2+2x+1)+4Cx^3-Cx+4Dx^2-D)/((2x+1)(2x-1)(4x^2+1))
x/(16x^4-1)=(x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D)/((2x+1)(2x-1)(4x^2+1))
:.x=x^3(8A+8B+4C)+x^2(-4A+4B+4D)+x(2A+2B-C)-A+B-D
equating the coefficients of the like terms,
8A+8B+4C=0 ----- equation 1
-4A+4B+4D=0 ----- equation 2
2A+2B-C=1 ----- equation 3
-A+B-D=0 ------ equation 4
Now we have to solve the above four equations to find the solutions of A,B,C and D.
From equation 1,
4(2A+2B+C)=0
2A+2B+C=0
Subtract equation 3 from the above equation,
(2A+2B+C)-(2A+2B-C)=0-1
2C=-1
C=-1/2
From equation 2,
4(-A+B+D)=0
-A+B+D=0
Now subtract equation 4 from the above equation,
(-A+B+D)-(-A+B-D)=0
2D=0
D=0
Now plug in the values of C in the equation 3,
2A+2B-(-1/2)=1
2A+2B+1/2=1
2A+2B=1-1/2
2(A+B)=1/2
A+B=1/4 ----- equation 5
Plug in the value of D in the equation 4,
-A+B-0=0
-A+B=0 ---- equation 6
Now add the equations 5 and 6,
2B=1/4
B=1/8
Plug in the value of B in the equation 6,
-A+1/8-0
A=1/8
:.x/(16x^4-1)=(1/8)/(2x+1)+(1/8)/(2x-1)+((-1/2)x)/(4x^2+1)
x/(16x^4-1)=1/(8(2x+1))+1/(8(2x-1))-x/(2(4x^2+1))