#I=intdx/((x-3)sqrt(6x-x^2-5))#
Complete the square in the square root:
#I=intdx/((x-3)sqrt(-(x^2-6x+9)-5+9))#
#I=intdx/((x-3)sqrt(4-(x-3)^2))#
Let #x-3=t#. This implies that #dx=dt#:
#I=intdt/(tsqrt(4-t^2))#
Now let #t=2sintheta#. This implies that #dt=2costhetad theta#.
#I=int(2costhetad theta)/(2sinthetasqrt(4-4sin^2theta))#
Note that #sqrt(4-4sin^2theta)=2sqrt(1-sin^2theta)=2costheta#:
#I=int(2costhetad theta)/(2sintheta(2costheta))#
#I=1/2intcscthetad theta#
This is a well known integral:
#I=-1/2lnabs(csctheta+cottheta)#
#I=-1/2lnabs((1+costheta)/sintheta)#
#I=1/2lnabs(sintheta/(1+sqrt(1-sin^2theta)))#
Our substitution #t=2sintheta# implies that #sintheta=t//2#:
#I=1/2lnabs((t//2)/(1+sqrt(1-(t//2)^2)))#
Note that #sqrt(1-(t//2)^2)=sqrt((4-t^2)/4)=1/2sqrt(4-t^2)#:
#I=1/2lnabs((t//2)/(1+1/2sqrt(4-t^2)))#
#I=1/2lnabs(t/(2+sqrt(4-t^2)))#
Using #t=x-3#:
#I=1/2lnabs((x-3)/(2+sqrt(6x-x^2-5)))+C#
