The expression I'm trying to express using partial fractions is: $\frac{64x}{(2x-1)^2(2x+1)^2}$
What I've tried so far:
$\frac{64x}{(2x-1)^2(2x+1)^2}=\frac{A}{(2x-1)^2}+\frac{B}{2x-1}+\frac{C}{(2x+1)^2}+\frac{D}{2x+1}$ where $A,B,C,D$ are constants
Multiplying each term by $(2x-1)^2(2x+1)^2$ gives
$64x=A(2x+1)^2+B(2x-1)(2x+1)^2+C(2x-1)^2+D(2x+1)(2x-1)^2$
Now:
$x=-\frac{1}{2} \implies -32=4C \implies C=-8$
$x=\frac{1}{2} \implies 32=4A \implies A=8$
$x=0 \implies 0=A-B+C+D \implies B=D$
$x=1 \implies 64=9A+9B+C+3D \implies 0=3B+D$
Solving the two simultaneous equations yields $B=D=0$
Hence: $\frac{64x}{(2x-1)^2(2x+1)^2}=\frac{8}{(2x-1)^2}-\frac{8}{(2x+1)^2}$
But we know this is wrong since, checking with $x=1$ gives:
$\frac{64}{9}=\frac{1}{9}$ which is obviously wrong.
So my questions are:
(1) What method(s) can we use to express this in partial fractions?
(2) What went wrong above?
I'm not sure how you get $\frac19$ when you checked for $x=1$. Notice that
$$\frac8{(2(1)-1)^2}=\frac8{(2-1)^2}=\frac81=8$$
$$\frac8{(2(1)+1)^2}=\frac8{(2+1)^2}=\frac89$$
Thus,
$$\frac8{(2(1)-1)^2}-\frac8{(2(1)+1)^2}=8-\frac89=\frac{64}9$$
Similarly,
$$\frac{64(1)}{(2(1)-1)^2(2(1)+1)^2}=\frac{64}{1^23^2}=\frac{64}9$$
Just as suspected.
As far as I can tell, your partial fractions are done correctly.