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

Question

Isabella Ackerman · Accepted Answer

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#

Emily Ammerman · Answer

undefined To integrate $ \frac{x}{16x^4 - 1} $ using partial fractions, first, factor the denominator $ 16x^4 - 1 $ into irreducible factors:

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

The partial fraction decomposition will have the form:

$ \frac{x}{16x^4 - 1} = \frac{A}{2x - 1} + \frac{B}{2x + 1} + \frac{Cx + D}{2x^2 + 1} $

Solve for the constants $ A, B, C, $ and $ D $ by equating coefficients:

$ x = A(2x + 1)(2x^2 + 1) + B(2x - 1)(2x^2 + 1) + (Cx + D)(2x^2 - 1) $

$ A = \frac{1}{8} $, $ B = -\frac{1}{8} $, $ C = 0 $, $ D = \frac{1}{4} $

Now integrate each term:

$ \int \frac{x}{16x^4 - 1} \,dx = \frac{1}{8} \ln|2x - 1| - \frac{1}{8} \ln|2x + 1| + \frac{1}{4} \tan^{-1}(2x) + C $