## Calculus 8th Edition

$\frac{ln2}{4}$
$\int\int_{D} \frac{y}{1+x^{2}}dA=\int_{0}^{1}\int_{0} ^{\sqrt x} \frac{y}{1+x^{2}}dydx$ $=\int_{0}^{1}[\frac{y^{2}}{2}\cdot \frac{1}{1+x^{2}}]_{0} ^{\sqrt x\sqrt x}dx$ $=\frac {1}{4}\int_{0} ^{1}\frac{2x}{1+x^{2}}dx$ Put $u=1+x^{2}$ and $2xdx=du$ $=\frac{1}{4}\int_{1} ^{2}\frac{du}{u}$ $=\frac{1}{4}[lnu]_{1} ^{2}$ $=\frac{ln2}{4}$