`int x/(16x^4-1)dx`

To solve using partial fraction method, the denominator of the integrand should be factored.

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

Take note that if the factors in the denominator are linear, each factor has a partial fraction in the form `A/(ax+b)` .

If the factors are in quadratic form, each factor has a partial fraction in the form `(Ax+B)/(ax^2+bx+c)` .

So expressing the integrand as sum of fractions, it becomes:

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

To determine the values of A, B, C and D, multiply both sides by the LCD of the fractions present.

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

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

Then, assign values to x in which either `2x-1`, `2x+1`, or`4x^2+1` will become zero.

So plug-in `x=1/2` to get the value of A.

`1/2=A(2(1/2)+1)(4(1/2)^2+1)+B(2(1/2)-1)(4(1/2)^2 + 1) + (C(1/2)+D)(2(1/2)-1)(2(1/2)+1)`

`1/2=A(4) + B(0)+(C(1/2)+D)(0)`

`1/2=4A`

`1/8=A`

Plug-in `x=-1/2` to get the value of B.

`-1/2=A(2(-1/2)+1)(4(-1/2)^2+1) + B(2(-1/2)-1)(4(-1/2)^2+1) + (C(-1/2)+D)(2(-1/2)-1)(2(-1/2)+1)`

`-1/2=A(0)+B(-4) +(C(-1/2)+D)(0)`

`-1/2=-4B`

`1/8=B`

To solve for D, plug-in the values of A and B. Also, plug-in x=0.

`0=1/8(2(0)+1)(4(0)^2+1) + 1/8(2(0)-1)(4(0)^2+1) + (C(0)+D)(2(0)-1)(2(0)+1)`

`0=1/8 - 1/8 -D`

`0=D`

To solve for C, plug-in the values of A, B and D. Also, assign any value to x. Let it be x=1.

`1=1/8(2(1)+1)(4(1)^2+1) +1/8(2(1) -1)(4(1)^2+1) + (C(1) + 0)(2(1) -1)(2(1)+1)`

`1=15/8+5/8+C(3)`

`1=10/4+3C`

`-3/2=3C`

`-1/2=C`

So the partial fraction decomposition of the integrand is:

`int x/(16x^4-1)dx`

`=int(x/((2x-1)(2x+1)(4x^2+1)))dx`

`=int (1/(8(2x-1)) + 1/(8(2x+1)) - x/(2(4x^2+1)) )dx`

Then, express it as three integrals.

`=int 1/(8(2x-1))dx + int 1/(8(2x+1))dx - int x/(2(4x^2+1)) dx`

`=1/8int 1/(2x-1)dx + 1/8int 1/(2x+1)dx - 1/2int x/(4x^2+1) dx`

To take the integral of each, apply substitution method.

For the first integral, let the substitution be:

`u=2x-1`

`du=2dx`

`(du)/2=dx`

For the second integral, let the substitution be:

`v = 2x+1`

`dv=2dx`

`(dv)/2=dx`

And for the third integral, let the substitution be:

`w=4x^2+1`

`dw=8xdx`

`(dw)/8=xdx`

So the three integrals become:

`= 1/8 int 1/u * (du)/2 + 1/8 int 1/v *(dv)/2 -1/2 int 1/w * (dw)/8`

`=1/16 int 1/u du + 1/16 int 1/v dv - 1/16 int 1/w dx`

Then, apply the formula `int 1/x dx = ln|x| + C` .

`=1/16 ln|u| + 1/16 ln|v| - 1/16ln|w|+C`

And substitute back `u = 2x - 1` , `v = 2x + 1` and `w = 4x^2+1` .

`= 1/16ln|2x-1| + 1/16ln|2x+1| -1/16|ln4x^2+1|+C`

**Therefore, `int x/(16x^4-1)dx=1/16ln|2x-1| + 1/16ln|2x+1|
-1/16|ln4x^2+1|+C` .**

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