The answer is: #1/3ln(e^(3x)+sqrt(e^(6x)-36))+c#.
First of all I assume that there is an error in your writing: I think that #e^6x# would be #e^(6x)#.
Let's assume:
#e^(3x)=6coshtrArr3x=ln6coshtrArrx=1/3ln6coshtrArr#
#dx=1/3*(6sinht)/(6cosht)dt=1/3sinht/coshtdt#.
Our integral becomes:
#int(6cosht)/sqrt(36cosh^2t-36)*1/3sinht/coshtdt=2intsinht/(6sqrt(cosh^2t-1))dt=#
#=1/3intsinht/sinhtdt=1/3intdt=1/3t+c=(1)#.
Since #e^(3x)=6coshtrArrcosht=e^(3x)/6rArrt=arccosh(e^(3x)/6)#.
So:
#(1)=1/3arccosh(e^(3x)/6)+c#.
There is another way to write the solution, remembering the logarithmic expression of the function #y=arccoshx#, that is:
#y=ln(x+sqrt(x^2-1))#.
So:
#(1)=1/3ln(e^(3x)/6+sqrt(e^(6x)/36-1))+c=#
#=1/3ln((e^(3x)+sqrt(e^(6x)-36))/6)+c=#
#=1/3ln(e^(3x)+sqrt(e^(6x)-36))-1/3ln6+c=#
#=1/3ln(e^(3x)+sqrt(e^(6x)-36))+c#
because #-1/3ln6# is a number.
