Answer
$\dfrac{\ln 17}{4}$
Work Step by Step
We reverse the order of integration to get:
$I= \int_0^{2} \int_0^{y^3} \dfrac{1}{y^4+1} dx dy$
or, $= \int_0^{2} \dfrac{y^3}{y^4+1} dy$
Set $a =y^4+1 \implies da=4y^3 dy$
Now, $I=\dfrac{1}{4} \int_{1}^{17}\dfrac{da}{a}=\dfrac{\ln 17}{4}$
