Try letting #x=sectheta#. Note this implies that #dx=secthetatanthetad theta#.
Also, the bounds will change. Observe that #x=sqrt2=>theta=pi/4# and #x=2=>x=pi/3#.
Then:
#int_sqrt2^2dx/(x^5sqrt(x^2-1))=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetasqrt(sec^2theta-1))#
Recall the trigonometric identity #sec^2theta-1=tan^2theta#.
#=int_(pi/4)^(pi/3)(secthetatanthetad theta)/(sec^5thetatantheta)#
#=int_(pi/4)^(pi/3)(d theta)/sec^4theta#
#=int_(pi/4)^(pi/3)cos^4thetad theta#
From here, I would use the identity #cos^2alpha=1/2(1+cos2alpha)#, or, in this case, #cos^4alpha=1/4(1+cos2alpha)^2# to further simplify the integral. It will definitely be a little messy.
