Answer
The area of the surface obtained by rotating the curve about the x-axis is:
$$
S=\int_{0}^{1} 2 \pi y d s =\pi (e+1)
$$
Work Step by Step
$$
y= \sqrt {1+e^{x}}, \quad \quad 0 \leq x \leq 1
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}= \frac{1}{2}(1+e^{x})^{-\frac{1}{2}}. e^{x}=\frac{e^{x}}{2\sqrt {1+e^{x}}}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(\frac{e^{x}}{2\sqrt {1+e^{x}}} \right)^{2}} dx\\
&=\sqrt{1+\frac{e^{2 x}}{4\left(1+e^{x}\right)}}dx \\
&=\sqrt{\frac{4+4 e^{x}+e^{2 x}}{4\left(1+e^{x}\right)}} dx\\ &=\sqrt{\frac{\left(e^{x}+2\right)^{2}}{4\left(1+e^{x}\right)}}dx \\ &=\frac{e^{x}+2}{2 \sqrt{1+e^{x}}}dx.
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis is:
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{0}^{1} 2 \pi \sqrt {1+e^{x}} \frac{e^{x}+2}{2 \sqrt{1+e^{x}}}dx\\
&=\pi \int_{0}^{1} (e^{x}+2)dx\\
&=\pi [e^{x}+2x]_0^1\\
&=\pi [(e+2)-(1+0)]\\
&=\pi (e+1)
\end{aligned}
$$
