Whenever we encounter a square root of the form sqrt(x^2 - a), then this calls for substituting sec t, and subsequently making use of the trigonometric identity sec^2 t - 1 = tan^2 t. Taking this line, we substitute x \mapsto g(t) with
g(t) = 2 sec t,
g'(t) = 2 sec t tan t, and
g^{-1}(x) = sec^{-1} ( x / 2):
int \frac{1}{sqrt(x^2 - 4)^3} \ dx =
= [ \frac{1}{4} int \frac{sec t tan t}{sqrt(sec^2 t - 1)^3} \ dt ]_{t = sec^{-1} ( x / 2 )} =
= [ \frac{1}{4} int \frac{sec t}{tan^2 t} \ dt ]_{t = sec^{-1} ( x / 2 )} =
= [ \frac{1}{4} int \frac{cos t}{sin^2 t} \ dt ]_{t = sec^{-1} ( x / 2 ).
At this point, we may interpret sin t as an inner function (with derivative cos t), and can thus write the integral as
[ \frac{1}{4} int \frac{1}{u^2} \ du ]_{u = sin t = sin sec^{-1} ( x / 2 )}.
Remembering that
sin x = sqrt(sec^2 x - 1) / sec x,
we simplify the resubstitution, and get
[ \frac{1}{4} int \frac{1}{u^2} \ du ]_{u = sqrt(x^2 / 4 - 1) / (x/2) = sqrt(x^2 - 4) / x} =
[ - \frac{1}{4 u} + C ]_{u = sqrt(x^2 - 4) / x} =
- \frac{x}{4 sqrt(x^2 - 4)} + C.