The surface area is given by
[tex]\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx[/tex]
since [tex]y=1+e^x\implies y'=e^x[/tex]. To compute the integral, first let
[tex]u=e^x\implies x=\ln u[/tex]
so that [tex]\mathrm dx=\frac{\mathrm du}u[/tex], and the integral becomes
[tex]\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du[/tex]
[tex]=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du[/tex]
Next, let
[tex]u=\tan t\implies t=\tan^{-1}u[/tex]
so that [tex]\mathrm du=\sec^2t\,\mathrm dt[/tex]. Then
[tex]1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t[/tex]
so the integral becomes
[tex]\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt[/tex]
[tex]=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt[/tex]
Rewrite the integrand with
[tex]\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}[/tex]
so that integrating the first term boils down to
[tex]\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds[/tex]
where we substitute [tex]s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt[/tex]. Since
[tex]\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)[/tex]
the first term in this integral contributes
[tex]\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}[/tex]
[tex]=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}[/tex]
The second term of the integral contributes
[tex]\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt[/tex]
The antiderivative of [tex]\sec^3t[/tex] is well-known (enough that I won't derive it here myself):
[tex]\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C[/tex]
so this latter integral's contribution is
[tex]\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)[/tex]
Then the surface area is
[tex]2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)[/tex]
[tex]=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}[/tex]
