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.
