Find \( \iint_{R} \frac{x y^{2}}{x^{2}+1} d A, R=[0,2] \times[-3,3] \) Round your answer to four decimal places.

To solve the double integral \( \iint_{R} \frac{x y^{2}}{x^{2}+1} \, dA \) over the region \( R = [0,2] \times [-3,3] \), we can express it in terms of iterated integrals: \[ \iint_{R} \frac{x y^{2}}{x^{2}+1} \, dA = \int_{0}^{2} \int_{-3}^{3} \frac{x y^{2}}{x^{2}+1} \, dy \, dx. \] Let's first evaluate the inner integral: \[ \int_{-3}^{3} \frac{x y^{2}}{x^{2}+1} \, dy. \] Since \( \frac{x}{x^2+1} \) is constant with respect to \( y \), we can factor it out: \[ = \frac{x}{x^2 + 1} \int_{-3}^{3} y^{2} \, dy. \] Next, we calculate \( \int_{-3}^{3} y^{2} \, dy \): \[ \int_{-3}^{3} y^{2} \, dy = \left[ \frac{y^3}{3} \right]_{-3}^{3} = \frac{3^3}{3} - \frac{(-3)^3}{3} = \frac{27}{3} - \frac{-27}{3} = 9 + 9 = 18. \] Now substitute this back into the integral: \[ \int_{-3}^{3} \frac{x y^{2}}{x^{2}+1} \, dy = \frac{x}{x^{2} + 1} \cdot 18 = \frac{18x}{x^{2}+1}. \] Now substitute this result into the outer integral: \[ \int_{0}^{2} \frac{18x}{x^{2}+1} \, dx. \] To evaluate this integral, we can use the substitution \( u = x^2 + 1 \), so \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \). When \( x = 0 \), \( u = 1\), and when \( x = 2\), \( u = 5\): \[ \int_{0}^{2} \frac{18x}{x^{2}+1} \, dx = 18 \cdot \int_{1}^{5} \frac{1}{u} \cdot \frac{du}{2} = 9 \int_{1}^{5} \frac{du}{u}. \] Evaluating the integral \( \int \frac{du}{u} \): \[ 9 \int_{1}^{5} \frac{du}{u} = 9 \left[ \ln |u| \right]_{1}^{5} = 9 (\ln 5 - \ln 1) = 9 \ln 5. \] Calculating \( \ln 5 \): Using the approximation \( \ln 5 \approx 1.6094 \): \[ 9 \ln 5 \approx 9 \cdot 1.6094 = 14.4846. \] Now, rounding to four decimal places, we get: \[ \boxed{14.4846}. \]