$\int \frac{1}{(x^2-4)^2}dx$ 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 $$x=2\sec t$,$
but It caused some difficulty. 
 A: Perhaps the neatest approach is to square $\frac{1}{x^2-4}=\frac14\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$ to give$$\frac{1}{(x^2-4)^2}=\frac{1}{16}\left(\frac{1}{(x-2)^2}+\frac{1}{(x+2)^2}-\frac{2}{x^2-4}\right)=\frac{1}{32}\left(\frac{2}{(x-2)^2}+\frac{2}{(x+2)^2}+\frac{1}{x+2}-\frac{1}{x-2}\right).$$Now you can integrate:$$\int\frac{dx}{(x^2-4)^2}=\frac{1}{32}\left(-\frac{2}{x-2}-\frac{2}{x+2}+\ln\left|\frac{x+2}{x-2}\right|\right)+C,$$where $C$ is locally constant and can change at each of $x=\pm2$.
A: HINT:
$x=2\text{sec}(\theta)$, $dx=2\text{sec}(\theta)\text{tan}(\theta)d\theta$ then:
$\int \frac{1}{(x^2-4)^2}dx=\int \frac{2\text{sec}(\theta)\text{tan}(\theta)}{(4\text{sec}^2(\theta)-4)^2}d\theta=\frac{1}{8}\int \frac{\text{sec}(\theta)\text{tan}(\theta)}{\text{tan}^4(\theta)}d\theta$
$$=\frac{1}{8}\int \text{ctg}^2(\theta)\text{csc}(\theta) d\theta= \frac{1}{8}(\int \text{csc}^3(\theta)d\theta-\int\text{csc}(\theta) d\theta)$$
A: Hint:
Another way
Write numerator as $$\left(\dfrac{x+2-(x-2)}4\right)^2$$
Now $\dfrac1{(x+2)(x-2)}=\dfrac{x+2-(x-2)}{4(x+2)(x-2)}=?$
A: It is $$\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}
$$
A: Here is a compact approach with the substitution $x=2\cosh t$,
$$I= \int \frac{1}{(x^2-4)^2}dx$$
$$=\frac{1}{8} \int \text{csch}^3 tdt
= -\frac{1}{8} \int \text{csch} t \>d(\coth t)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \int \text{csch} t \coth^2 tdt \right)$$
$$= -\frac{1}{8} \left(\text{csch} t \coth t + \int \text{csch} t dt\right)-I$$
$$=-\frac{1}{16}\left( \text{csch} t \coth t + \int \text{csch} t dt\right)$$
$$=-\frac{1}{16} \left( \text{csch} t \coth t+ \ln \tanh\frac t2 \right)+C$$
