Answer
\[\frac{\sqrt{17}-\sqrt{2}}{2}\]
Work Step by Step
\[\begin{align}
& \iint\limits_{R}{\frac{2y}{\sqrt{{{x}^{4}}+1}}dA} \\
& \text{The region }R\text{ is given by} \\
& R=\left\{ \left( x,y \right):0\le y\le {{x}^{3/2}},\text{ }0\le x\le 1\text{ } \right\} \\
& \text{Then,} \\
& \iint\limits_{R}{\frac{2y}{\sqrt{{{x}^{4}}+1}}dA}=\int_{1}^{2}{\int_{0}^{{{x}^{3/2}}}{\frac{2y}{\sqrt{{{x}^{4}}+1}}}dy}dx \\
& \text{Integrating} \\
& =\int_{1}^{2}{\left[ \frac{{{y}^{2}}}{\sqrt{{{x}^{4}}+1}} \right]_{0}^{{{x}^{3/2}}}}dx \\
& =\int_{1}^{2}{\left[ \frac{{{\left( {{x}^{3/2}} \right)}^{2}}}{\sqrt{{{x}^{4}}+1}}-\frac{{{\left( 0 \right)}^{2}}}{\sqrt{{{x}^{4}}+1}} \right]}dx \\
& =\int_{1}^{2}{\frac{{{x}^{3}}}{\sqrt{{{x}^{4}}+1}}}dx \\
& =\frac{1}{4}\int_{1}^{2}{\frac{4{{x}^{3}}}{\sqrt{{{x}^{4}}+1}}}dx \\
& =\frac{1}{4}\left[ \frac{\sqrt{{{x}^{4}}+1}}{1/2} \right]_{1}^{2} \\
& =\frac{1}{2}\left[ \sqrt{17}-\sqrt{2} \right] \\
& =\frac{\sqrt{17}-\sqrt{2}}{2} \\
\end{align}\]
