Answer
$\dfrac{\ln (17)}{4}$
Work Step by Step
Consider $I=\int_{0}^{8} \int_{\sqrt[3] x}^{2} \dfrac{1}{y^4+1} \ dy \ dx$
or, $=\int_{0}^{2} \int_{0}^{y^3} \dfrac{1}{y^4+1} \ dx \ dy$
or, $=\dfrac{1}{4} \times \int_{0}^{2} \dfrac{4y^3}{y^4+1} \ dy$
Plu $y^4+1=u$ and $du=4y^3 dy$
Now, $I=(1/4) \int_{0}^2 u^{-1} du$
or, $=(1/4) [\ln u]_0^2$
or, $=\dfrac{\ln (17)}{4}$
