L=int_0^1sqrt{1+({dy}/{dx})^2} dxL=∫10√1+(dydx)2dx
=int_0^1sqrt{1+e^{2x}}dx=∫10√1+e2xdx
by the substitution u=sqrt{1+e^{2x}}u=√1+e2x.
Rightarrow {du}/{dx}=e^{2x}/sqrt{1+e^{2x}}={u^2-1}/u
Rightarrow dx={u}/{u^2-1}du⇒dudx=e2x√1+e2x=u2−1u⇒dx=uu2−1du
As xx goes from 00 to 11, uu goes from sqrt{2}√2 to sqrt{1+e^2}√1+e2
=int_{sqrt{2}}^{sqrt{1+e^2}}u^2/{u^2-1} du=∫√1+e2√2u2u2−1du
by partial fraction decomposition,
=int_{sqrt{2}}^{sqrt{1+e^2}}[1+1/2(1/{u-1}-1/{u+1})] du=∫√1+e2√2[1+12(1u−1−1u+1)]du
=[u+1/2(ln|u-1|-ln|u+1|)]_{sqrt{2}}^{sqrt{1+e^2}}=[u+12(ln|u−1|−ln|u+1|)]√1+e2√2
=[u+1/2ln|{u-1}/{u+1}|]_{sqrt{2}}^{sqrt{1+e^2}}=[u+12ln∣∣∣u−1u+1∣∣∣]√1+e2√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})=√1+e2+12ln(√1+e2−1√1+e2−1)−√2−12ln(√2−1√2+1)
I hope that this was helpful.
