Answer
$9\ln 2$
Work Step by Step
$\displaystyle \iint_{R}\frac{xy^{2}}{x^{2}+1}dA=\int_{0}^{1}\int_{-3}^{3}\frac{xy^{2}}{x^{2}+1}dydx=$
Let $ g(x)=\displaystyle \frac{x}{x^{2}+1},\quad h(y)=y^{2}\quad$ and apply formula (5),
$\displaystyle \iint_{R}g(x)\cdot h(y)dA=\int_{a}^{b}g(x)dx\cdot\int_{c}^{d}h(y)dy$
$=\displaystyle \int_{0}^{1}\frac{x}{x^{2}+1}dx\cdot\int_{-3}^{3}y^{2}dy$
... for the first integral, $\left[\begin{array}{ll}
t=x^{2}+1 & \\
dt=2xdx & \\
x=0\Rightarrow t=1 & x=1\Rightarrow t=2
\end{array}\right]$
$=\displaystyle \frac{1}{2}\int_{1}^{2}t^{-1}dt\cdot\int_{-3}^{3}y^{2}dy$
$=[\displaystyle \frac{1}{2}\ln|t|]_{1}^{2}\cdot[\frac{y^{3}}{3}]_{-3}^{3}$
$=\displaystyle \frac{1}{2}(\ln 2-\ln 1)\cdot\frac{27-(-27)}{3}$
$=\displaystyle \frac{2\cdot 27}{2\cdot 3}(\ln 2-0)$
$=9\ln 2$
