How do you use partial fractions to find the integral #int x/(16x^4-1)dx#?

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Answer 1 · 2017-07-24T09:35:40

The answer is #=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C#

We need

#intdx/x=ln(|x|)+C#

Let's factorise the denominator

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

We can perform the decomposition into partial fractions

#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+))#

The denominators are the same, we compare the numerators

#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=-4A#, #=>#, #A=1/8#

Let #x=1/2#, #=>#, #1/2=4B#, #=>#, #B=1/8#

Let #x=0#, #=>#, #0=-A+B-D#, #=>#, #D=0#

Coefficients of #x^3#,

#0=8A+8B+4C#, #=>#, #4C=-2#, #C=-1/2#

Therefore,

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

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

#=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C#