For #x in (1,+oo)# substitute:
#x = sect#
#dx = sect tant dt#
with #t in (0,pi/2)#
so:
#int (2dx)/(x^3sqrt(x^2-1)) = 2 int (sect tant dt )/(sec^3tsqrt(sec^2t-1))#
Now:
#sec^2t -1 = tan^2t#
and as for #t in (0,pi/2)# the tangent is positive:
#sqrt(sec^2t -1) = tant#
then:
#int (2dx)/(x^3sqrt(x^2-1)) = 2 int (sect tant dt )/(sec^3t tant)#
#int (2dx)/(x^3sqrt(x^2-1)) = 2 int dt/sec^2t#
#int (2dx)/(x^3sqrt(x^2-1)) = 2 int cos^2tdt#
Now:
#2cos^2t = 1+cos2t#
so:
#int (2dx)/(x^3sqrt(x^2-1)) = int (1+cos2t)dt#
and using linearity:
#int (2dx)/(x^3sqrt(x^2-1)) = int dt +int cos2tdt#
#int (2dx)/(x^3sqrt(x^2-1)) =t + 1/2sin2t+C#
To undo the substitution note that:
#x = sect => tant = sqrt(x^2-1)#
so:
#t = arctan(sqrt(x^2-1))#
and using the parametric fomulas:
#sin 2t = (2tant)/(1+tan^2t) = (2sqrt(x^2-1))/(1+x^2-1) = (2sqrt(x^2-1))/x^2#
So:
#int (2dx)/(x^3sqrt(x^2-1)) = arctan(sqrt(x^2-1)) + sqrt(x^2-1)/x^2+C#
By differentiating we can see that the solution is valid also for #x in (-oo,-1)#
