$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$ 
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty. 

It is $$\frac{1}{(x^2-4)^2}=1/16\, \left( x-2 \right) ^{-2}-1/32\, \left( x-2 \right) ^{-1}+1/16\,
 \left( x+2 \right) ^{-2}+1/32\, \left( x+2 \right) ^{-1}
$$
