How do you integrate #int 1/sqrt(e^(2x)+12e^x-13)dx# using trigonometric substitution?

Here,

#I=int1/sqrt(e^(2x)+12e^x-13)dx#

#=int1/sqrt(e^(2x)+12e^x+36-49)dx#

#=int1/sqrt((e^x+6)^2-7^2)dx#

Let,

#color(blue)(e^x+6=7secu=>e^x=7secu-6=>e^xdx=7secutanudu#

#color(blue)(=>dx=(7secutanudu)/(7secu-6)&,secu= (e^x+6)/7=>color(red)(cosu=7/(e^x+6)#

#:.I=int1/sqrt(7^2sec^2u-7^2)xx(7secutanu)/(7secu-6)du#

#=int1/(7tanu)xx(7secutanu)/(7secu-6)du#

#=intsecu/(7secu-6)du#

#=int1/(7-6cosu)du#

Now,subst. #color(violet)(tan(u/2)=t=>sec^2(u/2)*1/2du=dt#

#color(violet)(=>du=(2dt)/(1+tan^2(u/2))=(2dt)/(1+t^2) and cosx=(1- t^2)/(1+t^2)#

#:.I=int1/(7-6((1-t^2)/(1+t^2)))xx2/(1+t^2)dt=2intdt/(7+7t^2- 6+6t^2)#

#=2int1/(13t^2+1)dt=2/13int1/(t^2+(1/13))dt#

#=2/13int1/(t^2+(1/sqrt13)^2)dt#

#=2/13xx1/(1/sqrt13)tan^-1(t/(1/sqrt13))+c#

#=2/sqrt13tan^-1(sqrt13t)+c, where,color(violet)(t=tan(u/2)#

#:.I=2/sqrt13tan^-1(sqrt13 tan(u/2))+c...to(A)#

We know that, #tan^2(u/2)=(1-cosu)/(1+cosu),where, color(red)(cosu=7/(e^x+6)#

#=>tan^2(u/2)=(1-(7/(e^x+6)))/(1+(7/(e^x+6)))=(e^x+6- 7)/(e^x+6+7)=(e^x-1)/(e^x+13)#

#=>tan(u/2)=sqrt((e^x-1)/(e^x+13))#

Hence, from #(A)#, we have

#I=2/sqrt13tan^-1(sqrt13xxsqrt((e^x-1)/(e^x+13)))+c#