How do you integrate #int 1/sqrt(x^2-9x+8) # using trigonometric substitution?

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)#

#=> int 1/ ( sqrt(( x-9/2)^2 -49/4 ) #

#=> u = (2x-9) / 7 #

#=> dx = 7/2 du #

#=> int 7 / sqrt( 49u^2 - 49 ) dx #

#=> int 1/ sqrt(u^2 -1 ) du #

#=> u = cosh theta #

#=> du = sinh theta d theta #

#=> int (sinh theta d theta ) /sqrt(cosh^2 theta -1) #

#=> int d theta #

#=> theta + c #

#=> cosh ^(-1)u + c #

#=> cosh^(-1) ((2x-9)/7 ) + c #