First, factorize the denominator to get #(4x^2+1)(2x+1)(2x-1)#.
We need to find #A#, #B#, #C#, and #D# such that #x/(16x^4-1)=(Ax+B)/(4x^2+1)+C/(2x+1)+D/(2x-1)# for all #x#. Multiply both sides by #(4x^2+1)(2x+1)(2x-1)#.
Then, #x=(Ax+B)(2x+1)(2x-1)+C(4x^2+1)(2x-1)+D(4x^2+1)(2x+1)#.
Set #x=1/2# to get #1/2=D*2*2#, or #D=1/8#.
Set #x=-1/2# to get #-1/2=C*2*-2#, or #C=1/8#.
Set #x=i/2# to get #i/2=((Ai)/2+B)(i+1)(i-1)=(A(i/2)+B)*-2#, or #i=-2Ai-4B#. Thus, #A=-1/2# and #B=0#.
From the above, it can be seen that #x/(16x^4-1)=(-1/2x)/(4x^2+1)+(1/8)/(2x+1)+(1/8)/(2x-1)#.
The problem the becomes #int\ ((-1/2x)/(4x^2+1)+(1/8)/(2x+1)+(1/8)/(2x-1))\ dx#, or #-1/2int\ x/(4x^2+1)\ dx+1/8int\ 1/(2x+1)\ dx+1/8int\ 1/(2x-1)\ dx#.
For the first integral, use the substitution #u=4x^2+1# and #du=8x\ dx# to get #-1/2int\ x/(4x^2+1)\ dx=-1/16int\ 1/u\ du=-1/16ln|u|+C#. Substitute #u=4x^2+1# to get #-1/16ln|4x^2+1|+C#.
For the second integral, use the substitution #u=2x+1# and #du=2\ dx# to get #1/8int\ 1/(2x+1)\ dx=1/16int\ 1/u\ du=1/16ln|u|+C#. Substitute #u=2x+1# to get #1/16ln|2x+1|+C#.
For the third integral, use the substitution #u=2x-1# and #du=2\ dx# to get #1/8int\ 1/(2x-1)\ dx=1/16int\ 1/u\ du=1/16ln|u|+C#. Substitute #u=2x-1# to get #1/16ln|2x-1|+C#.
Combine these to get the final answer #-1/16ln|4x^2+1|+1/16ln|2x+1|+1/16ln|2x-1|+C=1/16ln|(4x^2-1)/(4x^2+1)|+C#.
