Answer
$$\dfrac{\ln 17}{4}$$
Work Step by Step
$$\int_{R} dA= \int_0^{2} \int_0^{y^3} \dfrac{1}{y^4+1} dx dy\\= \int_0^{2} \dfrac{y^3}{y^4+1} dy$$
Suppose $y^4+1=t$ and $4y^3 dy =dt$
Now, $$\dfrac{1}{4} \int_{1}^{17}\dfrac{dt}{t}=\dfrac{\ln 17}{4}$$
