We use #sec^2u=1+tan^2u#
Let's use the substitution
#x=(5secu)/3# #=>#, #dx=(5secutanudu)/3#
#intdx/(9x^2-25)^(3/2)=int((5secutanudu)/3)/(9*25/9sec^2u-25)^(3/2)#
#=int(5secutanudu)/(3*125(tan^3u))#
#=1/75int(secudu)/tan^2u#
#=1/75int(cos^2udu)/(cosu sin^2u)#
#=1/75int(cosudu)/(sin^2u)#
Let #v=sinu#,#=>##dv=cosudu#
Therefore,
#1/75int(cosudu)/(sin^2u)=1/75int(dv)/v^2#
#=1/75intv^(-2)dv=1/75v^(-1)/-1=-1/(75v)#
#=-1/(75sinu)#
#x=(5secu)/3# #=>#, #cosu=5/(3x)#
#sin^2u=1-cos^2u=1-25/(9x^2)=(9x^2-25)/(9x^2)#
#1/sinu=(3x)/(sqrt(9x^2-25))#
Therefore,
#intdx/(9x^2-25)^(3/2)=(-1/75)*(3x)/(sqrt(9x^2-25))+C#
#=-x/(25*sqrt(9x^2-25))+C#