How do you express #x/(16x^4-1)# in partial fractions?

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Answer 1 · 2016-11-20T08:57:38

The answer is #=(1/8)/(2x-1)+(1/8)/(2x+1)+(-x/2)/(4x^2+1)#

Let's factorise the denominator

#16x^4-1=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)#

We can start the decomposition into partial fractions.

So, #x/(16x^4-1)=x/((2x-1)(2x+1)(4x^2+1))#

#=A/(2x-1)+B/(2x+1)+(Cx+D)/(4x^2+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))#

so, #x=A(2x+1)(4x^2+1)+B(2x-1)(4x^2+1)+(Cx+D)(2x-1)(2x+1))#

let #x=-1/2#
#-1/2=B*-2*2#, #=># #B=1/8#

Let #x=1/2#
#1/2=A*2*2#, #=># #A=1/8#

let #x=0#
#0=A-B-D# #=>#, #D=0#

Coefficients of #x^3#
#0=8A+8B+4C#, #=>#, #C=-2/4=-1/2#

Therefore,

#x/(16x^4-1)=(1/8)/(2x-1)+(1/8)/(2x+1)+(-x/2)/(4x^2+1)#