Answer
$$\frac{4+\pi}{2} $$
Work Step by Step
Since
$$\int_{0}^{2} \frac{s+1}{\sqrt{4-s^{2}}} d s=\frac{1}{2} \int_{0}^{2} \frac{2 s d s}{\sqrt{4-s^{2}}}+\int_{0}^{2} \frac{d s}{\sqrt{4-s^{2}}}$$
For the first integral, let:
$u=4-s^2 \ \to \ du = -2sds$ at $s= 0,\to u= 4$ at $s= 2,\to u= 0$ , then
\begin{align*}
\int_{0}^{2} \frac{s+1}{\sqrt{4-s^{2}}} d s&=-\frac{1}{2} \int_{4}^{0} \frac{d u}{\sqrt{u}}+\lim _{c \rightarrow 2^{-}} \int_{0}^{c} \frac{d s}{\sqrt{4-s^{2}}}\\
&=\lim _{b \rightarrow 0^{+}} \int_{b}^{4} \frac{d u}{2 \sqrt{u}}+\lim _{c \rightarrow 2^{-}} \int_{0}^{c} \frac{d s}{\sqrt{4-s^{2}}}\\
&=\lim _{b \rightarrow 0^{+}}[\sqrt{u}]_{b}^{4}+\lim _{c \rightarrow 2^{-}}\left[\sin ^{-1} \frac{s}{2}\right]_{0}^{c}\\
&=\lim _{b \rightarrow 0^{+}}(2-\sqrt{b})+\lim _{c \rightarrow 2^{-}}\left(\sin ^{-1} \frac{c}{2}-\sin ^{-1} 0\right)\\
&=(2-0)+\left(\frac{\pi}{2}-0\right)\\
&=\frac{4+\pi}{2}
\end{align*}
