Answer
$=\frac{(\sqrt{5}-1) \pi}{4}$
Work Step by Step
$\int_{0}^{\sqrt{2}} \int_{y}^{\sqrt{4-y^{2}}} \frac{1}{\sqrt{x^{2}+y^{2}+1}} d x d y$
convert the limits to polar coordinates $R=\left\{(r, \theta): 0 \leqslant \theta \leqslant \frac{\pi}{4}, 0 \leqslant r \leqslant 2\right\}$
so
$=\int_{0}^{\frac{\pi}{4}} \int_{0}^{2} \frac{r}{\sqrt{r^{2}+1}} d r d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \int_{0}^{2}(\sqrt{r^{2}+1})^{-1 / 2}(2 r) d r d \theta$
integrate with respect to $r$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{4}}\left[\frac{(\sqrt{1+r^{2}})^{1 / 2}}{\frac{1}{2}}\right]_{0}^{2} d \theta$
$=\left(\int_{0}^{\frac{\pi}{4}}(\sqrt{1+2^{2}})^{1 / 2}-\left(1+0^{2}\right)^{1 / 2}\right) d \theta$
$=\int_{0}^{\frac{\pi}{4}}(-1+\sqrt{5}) d \theta$
evaluate and integrate $=(-1+\sqrt{5})[\theta]_{0}^{\pi / 4}$
$=\frac{(\sqrt{5}-1) \pi}{4}$
