The area is given by the integral,
[tex]\displaystyle\int_0^{\ln5}2\pi x(y)\sqrt{1+\left(\dfrac{\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy[/tex]
We have
[tex]x=\dfrac{e^y+e^{-y}}2\implies\dfrac{\mathrm dx}{\mathrm dy}=\dfrac{e^y-e^{-y}}2[/tex]
So now compute the integral:
[tex]\displaystyle\frac\pi2\int_0^{\ln5}(e^y+e^{-y})\sqrt{4+(e^y-e^{-y})^2}\,\mathrm dy[/tex]
Substitute [tex]u=e^y-e^{-y}[/tex] and [tex]\mathrm du=(e^y+e^{-y})\,\mathrm dy[/tex]:
[tex]\displaystyle\frac\pi2\int_0^{\frac{24}5}\sqrt{4+u^2}\,\mathrm du[/tex]
Another substitution, [tex]u=2\tan v[/tex] and [tex]\mathrm dv=2\sec^2v\,\mathrm dv[/tex]:
[tex]\displaystyle\frac\pi2\int_0^{\tan^{-1}\frac{12}5}\sqrt{4+(2\tan v)^2}\,2\sec^2v\,\mathrm dv[/tex]
[tex]\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sqrt{1+\tan^2v}\,\sec^2v\,\mathrm dv[/tex]
[tex]\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv[/tex]
There's a well-known formula for the integral of secant cubed, but if you don't know it off the top of your head (like me), you can integrate by parts:
[tex]\displaystyle I=\int\sec^3v\,\mathrm dv=\sec v\tan v-\int\sec v\tan^2v\,\mathrm dv[/tex]
Expand the remaining the integral in terms of powers of secant:
[tex]\displaystyle\int\sec v\tan^2v\,\mathrm dv=\int\sec v(\sec^2v-1)\,\mathrm dv=\int\sec^3v\,\mathrm dv-\int\sec v\,\mathrm dv[/tex]
so that
[tex]I=\sec v\tan v-\left(I-\displaystyle\int\sec v\,\mathrm dv\right)[/tex]
[tex]2I=\sec v\tan v+\displaystyle\int\sec v\,\mathrm dv[/tex]
[tex]\implies I=\displaystyle\int\sec^3v\,\mathrm dv=\frac{\sec v\tan v}2+\frac12\ln|\sec v+\tan v|+C[/tex]
Coming back to the area integral, we use the formula above to get
[tex]\displaystyle2\pi\int_0^{\tan^{-1}\frac{12}5}\sec^3v\,\mathrm dv=\pi\left(\sec v\tan v+\ln|\sec v+\tan v|\right)\bigg|_0^{\tan^{-1}\frac{12}5}[/tex]
Next,
[tex]\tan\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{12}5[/tex]
[tex]\sec\left(\tan^{-1}\dfrac{12}5\right)=\dfrac{13}5[/tex]
[tex]\tan0=0[/tex]
[tex]\sec0=1[/tex]
so the area is
[tex]\pi\left(\dfrac{13}5\cdot\dfrac{12}5+\ln\left(\dfrac{13}5+\dfrac{12}5\right)-1\cdot0-\ln(1+0)\right)=\boxed{\left(\dfrac{156}{25}+\ln5\right)\pi}[/tex]
