Answer
$4\ln \left(3\right)-4$
Work Step by Step
Given
$$ y =\frac{1}{2+\sqrt{x}},\ \ \ y =\frac{1}{2-\sqrt{x}},\ \ \ x=1 $$
First, we find the intersection points
\begin{aligned}
\frac{1}{2+\sqrt{x}}&=\frac{1}{2-\sqrt{x}}\\
2+\sqrt{x}&=2-\sqrt{x}\\
\sqrt{x}&=-\sqrt{x}
\end{aligned}
Then $x=0$ and $$ \frac{1}{2-\sqrt{x}}\geq \frac{1}{2+\sqrt{x}},\ \ \ 0\leq x\leq 1 $$
Hence
\begin{aligned}
A&= \int_{0}^{1} \left(\frac{1}{2-\sqrt{x}}-\frac{1}{2+\sqrt{x}}\right)dx\\
&= \int _0^1\frac{1}{2-\sqrt{x}}dx-\int _0^1\frac{1}{2+\sqrt{x}}dx
\end{aligned}
Let $u^2=x \ \to \ \ 2udu=dx$ and $x=0\ \to u=0 , x=1\ \ \to u=1 $
\begin{aligned}
A&= \int_{0}^{1} \left(\frac{1}{2-\sqrt{x}}-\frac{1}{2+\sqrt{x}}\right)dx\\
&=2 \int _0^1\frac{u}{2-u}du-2 \int _0^1\frac{u}{2+u}du\\
&= 2\left(\int _0^1\frac{2}{2-u}du-\int _0^11du\right)-2\left(\int _0^1du-\int _0^1\frac{2}{2+u}du\right) \\
&= 2\left(2\ln \left(2\right)-1\right)-2\left(1-2\left(\ln \left(3\right)-\ln \left(2\right)\right)\right)\\
&=4\ln \left(3\right)-4
\end{aligned}
