Answer
$$\frac{-1+\sqrt{17}}{2}$$
Work Step by Step
\[
\begin{array}{c}
\iint_{R} x\left(1+y^{2}\right)^{-1 / 2} d A=\int_{0}^{4}\left[\int_{0}^{\sqrt{y}} x\left(1+y^{2}\right)^{-1 / 2} d x\right] d y \\
=\int_{0}^{4}\left[\int_{0}^{\sqrt{y}} \frac{x^{2}}{2}\left(1+y^{2}\right)^{-1 / 2}\right] d y \\
=\int_{0}^{4} \frac{y}{2 \sqrt{y^{2}+1}} d y
\end{array}
\]
Relpace $2 y d y=d u$ and $u=y^{2}+1$
$=\int_{1}^{17} \frac{d u}{4 \sqrt{u}}$
$=\left[\frac{\sqrt{u}}{2}\right]_{1}^{17}$
\[
=\frac{-1+\sqrt{17}}{2}
\]
