Here,
#I=int1/sqrt(-e^(2x)+12e^x+72)dx#
Now,
#-e^(2x)+12e^x+72=color(brown)(108)-e^(2x)+12e^x-
color(brown)(36)=108-(e^x-6)^2#
#:.I=int 1/sqrt(108-(e^x-6)^2)dx#
Substitute ,#color(red)(e^x-6=sqrt108sinu)=>e^x=sqrt108
sinu+6#
#=>e^xdx#=#sqrt108cosudu=>dx=(sqrt108sinu)/(e^x)#=#
(sqrt108cosu)/(sqrt108sinu+6)#
So. #I=int1/sqrt(108-108sin^2u)*
(sqrt108cosu)/(sqrt108sinu+6)du#
#I=int1/(sqrt108cosu) xx (sqrt108cosu)/(sqrt108sinu+6)du#
#=int1/(sqrt108sinu+6)du#
#=1/6int1/(sqrt3sinu+1)du#
Substitute , #color(blue)(tan(u/2)=t)=>sec^2(u/2)1/2du=dt#
#=>du=(2dt)/(1+tan^2(u/2))=>color(blue)(du=(2dt)/(1+t^2)) and #
#sinu=(2tan(u/2))/(1+tan^2(u/2))=>color(blue)(sinu=(2t)/(1+t^2)#
So , #I=1/6int1/(sqrt3((2t)/(1+t^2))+1)*(2dt)/(1+t^2)#
#=>I=2/6int1/(2sqrt3t+1+t^2)dt#
#=>I=2/6int1/(t^2+2sqrt3t+3-2)dt to#[completing square]
#=>I=1/3int1/((t+sqrt3)^2-(sqrt2)^2)dt#
#=>I=1/3*1/(2sqrt2)int [((t+sqrt3+sqrt2)-(t+sqrt3-
sqrt2))/((t+sqrt3+sqrt2)(t+sqrt3-sqrt2))]dt#
#=1/(6sqrt2)int[1/(t+sqrt3-sqrt2)-1/(t+sqrt3+sqrt2)]dt#
#=1/(6sqrt2)[ln|t+sqrt3-sqrt2|-ln|t+sqrt3+sqrt2|]+c#
#:.I=1/(6sqrt2){ln|(t+sqrt3-sqrt2)/(t+sqrt3+sqrt2)|}+c...to(1)#
Now, #color(blue)(t=tan(u/2)#
#=>t^2=tan^2(u/2)=(1-cosu)/(1+cosu)=(1-cosu)/(1+cosu)
xxcolor(green)((1-cosu)/(1-cosu))#
#=>t^2=(1-cosu)^2/(1-cos^2u)=(1-cosu)^2/sin^2u#
#=>t=(1-cosu)/sinu=(1-sqrt(1-sin^2u))/sinu#
Subst. back #sinu=(e^x-6)/sqrt108#
#t=(1-sqrt(1-((e^x-6)/sqrt108)^2))/((e^x-6)/sqrt108)=>color(violet)(t=(sqrt108-sqrt(108-(e^x-6)^2))/(e^x-6))#
#:. t=(6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6)#
Hence ,from #(1)#
#I=1/(6sqrt2){ln|(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3-sqrt2)/(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3+sqrt2)|}+c#
#I#=#1/(6sqrt2)ln|(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3-sqrt2)(e^x-6))/(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3+sqrt2)(e^x-6))|##+c#