Here,
#I=int 1/sqrt(-e^(2x)-12e^x-35)dx#
Now,
#-e^(2x)-12e^x-35=1-(e^(2x)+12e^x+36)=1-(e^x+6)^2#
#:.I=int1/sqrt(1-(e^x+6)^2)dx#
Subst. #e^x+6=cosu=>e^x=cosu-6=>e^xdx=-sinudu#
#=>dx=-sinu/e^xdu=-sinu/(cosu-6)du=sinu/(6-cosu)du#
#I=int1/sqrt(1-cos^2u)xxsinu/(6-cosu)du#
#=int1/sinuxxsinu/(6-cosu)du#
#=int1/(6-cosu)du#
Again subst. #tan(u/2)=t=>sec^2(u/2)*1/2du=dt#
#=>du=(2dt)/(1+tan^2(u/2))=(2dt)/(1+t^2)and cosu=(1-
t^2)/(1+t^2)#
So,#I=int1/(6-(1-t^2)/(1+t^2))xx(2/(1+t^2))dt#
#=int2/(6+6t^2-1+t^2)dt#
#=int2/(7t^2+5)dt#
#=2/7int1/(t^2+(sqrt(5/7))^2)dt#
#=2/7xx1/(sqrt(5/7)) tan^-1(t/(sqrt(5/7)))+c#
#=2/sqrt35tan^-1((sqrt(7) t)/sqrt5)+c ,where, t=tan(u/2)#
#:.I=2/sqrt35tan^-1((sqrt(7) tan(u/2))/sqrt5)+c #
Now,
#tan^2(u/2)=(1-cosu)/(1+cosu),where, cosu=e^x+6#
#:.tan^2(u/2)=(1-(e^x+6))/(1+(e^x+6))=(-e^x-5)/(e^x+7) <0 ..??#
#=>tan(u/2)=sqrt((-e^x-5)/(e^x+7)#
Thus,
#I=2/sqrt35 tan^-1(sqrt(7(-e^x-5))/(sqrt(5(e^x+7))))+c#
