Answer
$\int_1^2 \int_1^2\dfrac{y}{x+y^2} \ dx \ dy=3 \ln (6) -\dfrac{3}{2} \ln (3)-\dfrac{5}{2} \ln (5) +\ln (2)$
Work Step by Step
Here, we have: $\int_1^2 \int_1^2\dfrac{y}{x+y^2} \ dx \ dy=\int_1^2 \int_1^2\dfrac{y}{x+y^2} \ dy \ dx$
or, $=\dfrac{1}{2} \times \int_1^2 [\ln (x+y^2)]_1^2 \ dx$
or, $=\dfrac{1}{2} \times \int_1^2 [\ln (x+4)-\ln (x+1)] \ dx$
or, $=\dfrac{1}{2} \times [(x+4) \ln (x+4) -x-(x+1) \ln (x+1)+x]_1^2 $
Thus, we get:
$\int_1^2 \int_1^2\dfrac{y}{x+y^2} \ dx \ dy=3 \ln (6) -\dfrac{3}{2} \ln (3)-\dfrac{5}{2} \ln (5) +\ln (2)$
