`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...
`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))`
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