Answer
$$\frac{1}{4}\ln(17)$$
Work Step by Step
Given $$\iint_{D} \frac{y}{x^{2}+1} d A, \quad D=\{(x, y) | 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant \sqrt{x}\}$$
From $D$, we have
\begin{align*}
\iint_{D} \frac{y}{x^{2}+1} d A&=\int_{0}^{4}\int_{0}^{\sqrt{x}}\frac{y}{x^{2}+1} dydx\\
&=\frac{1}{2}\int_{0}^{4} \frac{1}{x^{2}+1}[y^2]_{0}^{\sqrt{x}} dx\\
&=\frac{1}{2}\int_{0}^{4} \frac{x}{x^{2}+1} dx\\
&=\frac{1}{4}\ln (x^2+1)\bigg|_{0}^{4}\\
&=\frac{1}{4}\ln(17)
\end{align*}
