Answer
$6\sqrt{2}+\frac{-\ln \left(2\sqrt{2}+3\right)+\ln \left(-2\sqrt{2}+3\right)}{2}$
Work Step by Step
Given
$$y^2-x^2=1,\ \ \ \ y=3$$
First, we find the intersection points
\begin{aligned}
y^2-x^2&=1\\
9-x^2&= 1\\
x^2-8&=0
\end{aligned}
Then $x=\pm 2\sqrt{2}$, since
$$3\geq \sqrt{1+x^2},\ \ \ \ \ -2\sqrt{2}\leq x\leq 2\sqrt{2}$$
Hence
\begin{aligned}
A&=\int_{-2\sqrt{2}}^{2\sqrt{2} } \left(3-\sqrt{1+x^2}\right)dx \\
&= \int \:3dx-\int \sqrt{1+x^2}dx
\end{aligned}
Use the rule $$\int \sqrt{a^2+u^2} d u=\frac{u}{2} \sqrt{a^2+u^2}+\frac{a^2}{2} \ln \left(u+\sqrt{a^2+u^2}\right)+C$$
we get
\begin{aligned}
A&=\int_{-2\sqrt{2}}^{2\sqrt{2} } \left(3-\sqrt{1+x^2}\right)dx \\
&= \int \:3dx-\int \sqrt{1+x^2}dx\\
&= \left(3x-\frac{1}{2}x\sqrt{1+x^2}-\frac{1}{2}\ln \left|x+\sqrt{1+x^2}\right|\right)\bigg|_{-2\sqrt{2}}^{2\sqrt{2} }\\
&= 6\sqrt{2}+\frac{-\ln \left(2\sqrt{2}+3\right)+\ln \left(-2\sqrt{2}+3\right)}{2}
\end{aligned}
