Without Trig.Subst. answer ,see above.
With Trig. Subst. answer, see below.
Decide yourself ,which method is better.
We have, #x^2-9x+color(red)8=x^2-9x+color(red)(32/4)=x^2-
9x+color(red)(81/4-49/4)#
#:.x^2-9x+8=(x-9/2)^2-(7/2)^2#
#i.e.I=int1/sqrt((x-9/2)^2-(7/2)^2)dx#
Let, #x-9/2=7/2secu=>dx=7/2secutanudu#
So,
#I=int(7/2secutanu)/sqrt((7/2secu)^2-(7/2)^2)du#
#=int(7/2secutanu)/(7/2sqrt(tan^2u))du#
#=int(secutanu)/tanudu#
#=intsecudu#
#=ln|secu+tanu|+c#
#=ln|secu+sqrt(sec^2u-1)|+c,to[where,secu=(x-9/2)/(7/2)]#
#=ln|(x-9/2)/(7/2)+sqrt(((x-9/2)/(7/2))^2-1)|+c#
#=ln|(x-9/2)/(7/2)+sqrt((x-9/2)^2-(7/2)^2)/(7/2)|+c#
#=ln|(x-9/2)/(7/2)+sqrt(x^2-9x+8)/(7/2)|+c#
#=ln|(x-9/2+sqrt(x^2-9x+8))/(7/2)|+c#
#=ln|x-9/2+sqrt(x^2-9x+8)|-ln(7/2)+c#
#=ln|x-9/2+sqrt(x^2-9x+8)|+C,where,C=c-ln(7/2)#