## Thomas' Calculus 13th Edition

$$\dfrac{\ln (17)}{4}$$
We calculate the integral as follows: $$\int_{0}^{8} \int_{\sqrt[3] x}^{2} \dfrac{1}{y^4+1} \ dy \ dx =\int_{0}^{2} \int_{0}^{y^3} \dfrac{1}{y^4+1} \ dx \ dy \\= \dfrac{1}{4}\int_{0}^{2} \dfrac{4y^3}{y^4+1} \ dy$$ Set $y^4+1=a \implies 4y^3 dy = da$ Now, $$\dfrac{1}{4}\int_{0}^{2} \dfrac{4y^3}{y^4+1} dy= \int_{0}^2 \dfrac{a^{-1}}{4} da\\=(\dfrac{1}{4})\times [\ln u]_0^2 \\=\dfrac{\ln (17)}{4}$$