Here,
I=int1/sqrt(e^(2x)+12e^x-13)dxI=∫1√e2x+12ex−13dx
=int1/sqrt(e^(2x)+12e^x+36-49)dx=∫1√e2x+12ex+36−49dx
=int1/sqrt((e^x+6)^2-7^2)dx=∫1√(ex+6)2−72dx
Let,
color(blue)(e^x+6=7secu=>e^x=7secu-6=>e^xdx=7secutanuduex+6=7secu⇒ex=7secu−6⇒exdx=7secutanudu
color(blue)(=>dx=(7secutanudu)/(7secu-6)&,secu=
(e^x+6)/7=>color(red)(cosu=7/(e^x+6)⇒dx=7secutanudu7secu−6&,secu=ex+67⇒cosu=7ex+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
