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#.
