#L=int_0^1sqrt{1+({dy}/{dx})^2} dx#
#=int_0^1sqrt{1+e^{2x}}dx#
by the substitution #u=sqrt{1+e^{2x}}#.
#Rightarrow {du}/{dx}=e^{2x}/sqrt{1+e^{2x}}={u^2-1}/u
Rightarrow dx={u}/{u^2-1}du#
As #x# goes from #0# to #1#, #u# goes from #sqrt{2}# to #sqrt{1+e^2}#
#=int_{sqrt{2}}^{sqrt{1+e^2}}u^2/{u^2-1} du#
by partial fraction decomposition,
#=int_{sqrt{2}}^{sqrt{1+e^2}}[1+1/2(1/{u-1}-1/{u+1})] du#
#=[u+1/2(ln|u-1|-ln|u+1|)]_{sqrt{2}}^{sqrt{1+e^2}}#
#=[u+1/2ln|{u-1}/{u+1}|]_{sqrt{2}}^{sqrt{1+e^2}}#
#=sqrt{1+e^2}+1/2 ln({sqrt{1+e^2}-1}/{sqrt{1+e^2}-1})-sqrt{2}-1/2ln({sqrt{2}-1}/{sqrt{2}+1})#
I hope that this was helpful.
