How do you integrate #int 1/sqrt(4x^2-12x+34) # using trigonometric substitution?

Here,

#I=int1/sqrt(4x^2-12x+34)dx#

#=int1/sqrt(4x^2-12x+9+25)dx#

#=int1/sqrt((2x-3)^2+5^2)dx#

Subst, #2x-3=5tanu=>2dx=5sec^2udu#

#=>dx=5/2sec^2udu and color(blue)(tanu=(2x-3)/5#

So

#I=int1/sqrt(5^2tan^2u+5^2)xx5/2sec^2udu#

#=5/2intsec^2u/(5secu)du#

#=1/2intsecudu#

#=1/2ln|secu+tanu|+c#

#=1/2ln|sqrt(tan^2u+1)+tanu|+c#

Subst. back , #color(blue)(tanu=(2x-3)/5# we get

#I=1/2ln|sqrt(((2x-3)/5)^2+1)+((2x-3)/5)|+c#

#=1/2ln|(sqrt(4x^2-12x+34))/5+(2x-3)/5|+c#

#=1/2ln|((sqrt(4x^2-12x+34))+(2x-3))/5|+c#

#=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|-1/2ln5+c#

#=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|+C #

#where,C=-1/2ln5+c#

Here,

#I=int1/sqrt(4x^2-12x+34)dx#

#=int1/sqrt(4x^2-12x+9+25)dx#

#=int1/sqrt((2x-3)^2+5^2)dx#

Subst, #color(blue)(2x-3=u)=>2dx=du=>dx=1/2du#

So,

#I=int1/sqrt(u^2+5^2)xx1/2du#

#=1/2color(red)(int1/sqrt(u^2+5^2)du#

#=1/2color(red)(ln|u+sqrt(u^2+5^2)|+c#

Subst. back ,#color(blue)(u=2x-3 # ,we get

#I=1/2ln|(2x-3)+sqrt((2x-3)^2+5^2)|+c#

#=1/2ln|(2x-3)+sqrt(4x^2-12x+34)|+c#
.........................................................................................

Note :

#color(red)((1)int1/sqrt(x^2+a^2)dx=ln|x+sqrt(x^2+a^2)|+c#