Partial Fraction?

Question

x/(16x^4-1)
Need steps if avaliable also if there are any short cuts.

Answer 1

Since there are two stages of factoring, I think it is easier to do this in two steps. Find my approach here:
http://img329.imageshack.us/img329/7011/partialfractionbz4.png

Answer 2

x/(16x^4-1) =x / (x2 + 1)(x + 1)(x - 1)
Try to express this as
(ax+b)/(x2 + 1) + (cx + d)/(x+1) + (ex+f)/(x-1)
This must be true for all x.
Choose 6 different values of x (not being 1 or -1) and for that x demand that
(ax+b)/(x2 + 1) + (cx + d)/(x+1) + (ex+f)/(x-1) = x/(16x^4-1)
Then you get a set of 6 equations that you can solve.

Th

Answer 3

x/16x^4-1
=x/(4x^2+1)(2x+1)(2x-1)
=(Ax+B)/(4x^2+1)+C/(2x+1)
+D/(2x-1)
so (Ax+B)(4x^2-1)+C(4x^2+1)(2x-1)
+D(4x^2+1)(2x+1)=x
now remove the brackets and compare the coefficients to get A,B,C and D

Answer 4

x/(16x^4-1) can be factorized as
x/((4x^2-1)(4x^2+1)) which can be further factorized as
x/((2x-1)(2x+1)((4x^2+1))
now this can be splited into partial fractions as:

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

and rest I think u know how to solve

Answer 5

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

you could use trig subs, what is the context