Answer
$\frac{3}{2}ln 3-1$
Work Step by Step
$\int^{2}_{0} \int^{1}_{0} \frac{x}{1+xy} dx dy$
=$\int^{1}_{0} \int^{2}_{0} \frac{x}{1+xy} $
=$\int^{1}_{0}(ln(1+xy)]^{y=2}_{y=0})dx$
=$\int^{1}_{0} ln(1+2x)dx$
=$\frac{(1+2x)}{2}[ln(1+2x)-1]^1_0$
=$\frac{3}{2}ln3-1$
