Answer
$2 \ln 2$
Work Step by Step
Plug $u=1+t^2 \implies du=2x dx$
Re-arrange the given integral as follows:
$\int_{1}^{2} \int_{0}^{2}\dfrac{xy^3}{x^2+1} dx dy=(\dfrac{1}{2}) \int_{1}^{2} \int_{0}^{2}\dfrac{du}{u}y^3 dy$
This implies that
$(\dfrac{1}{2}) \int_{0}^{2} \ln 2]_{1}^{2} y^3 dy= (\dfrac{1}{2}) (\ln 2) (\dfrac{y^4}{4}]_{0}^{2}$
Hence, $(\dfrac{1}{2}) (\ln 2) (\dfrac{2^4}{4}]=2 \ln 2$
