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`