I need to evaluate the following integration problem: $\int_{0}^8 \int_{\sqrt[3]{y}}^2 \frac{y}{\sqrt{16+x^8}} dx dy$.
Since it cannot be evaluated in dx, the order of integeration must be changed as follow: $\int_{0}^2 \int_{0}^{x^3} \frac{y}{\sqrt{16+x^8}} dy dx$, and that can be reduced to: $\frac{1}{2}\int_{0}^2 \frac{x^6}{\sqrt{16+x^8}}dx$.
I've reached that point and couldn't move further. Is there a mistake in the problem itself (like the $x^8$ in the denumerator should be $x^7$ instead?)
tl;dr Given that the problem was found on a past exam, there was almost certainly a mistake in how the problem was printed; if the integrand had been $\frac{\mathbf 1}{\sqrt{16+x^8}}$, after swapping integration order and solving the easy first integral we get $\int_0^2\frac{x^3}{\sqrt{16+x^8}}\,dx$, which is elementary by the substitution $u=x^4$; the answer is $\frac{\sinh^{-1}4}4$. But for the answer to the question in the title, read on…
First substitute $u\sqrt2=x$ to transform the remaining single integral to $$I_0=\sqrt2\int_0^{\sqrt2}\frac{u^6}{\sqrt{1+u^8}}\,du$$ Now substitute $t=u^2+\frac1{u^2}$. By Byrd and Friedman 584.04 and .05 this reduces $I_0$ to elliptic integrals: $$I_0=\frac{\sqrt2}4\left(\int_2^\infty\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt+\int_2^{5/2}\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt+\int_{5/2}^\infty\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt\right)$$ The first two integrals above are solved by B&F 237.20 followed by 313.00, .02, .04: $$I_1(y)=\int_2^y\frac{t^2-t-1}{\sqrt{(t-2)(t^2-2)}}\,dt=\frac{\sqrt{4-2\sqrt2}}3\left[F(\varphi_1,m_1)-(2+\sqrt2)E(\varphi_1,m_1) + \sqrt{2+\sqrt2}\sqrt{\frac{(y-2)(y+\sqrt2)}{y-\sqrt2}}(y-\sqrt2+1)\right]$$ $$\varphi_1=\sin^{-1}\sqrt{\frac{y-2}{y-\sqrt2}},m_1=2\sqrt2-2$$ ($m$ is the parameter.) $I_1(\infty)$ has an infinite part of $\frac23\sqrt{\frac{(y-2)(y+\sqrt2)}{y-\sqrt2}}(y-\sqrt2+1)$; remember this for later. The last integral is solved by B&F 238.18 followed by 311.00, .02, .04: $$I_2(y)=\int_y^\infty\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt=\frac{\sqrt{4-2\sqrt2}}3\left[(1+\sqrt2)F(\varphi_2,m_2)-(2+\sqrt2)E(\varphi_2,m_2) + \sqrt{2-\sqrt2}\sqrt{\frac{y^2-2}{y+2}}(1+\sqrt2)(y+1)\right]$$ $$\varphi_2=\sin^{-1}\sqrt{\frac{\sqrt2+2}{y+2}},m_2=3-2\sqrt2$$ $$I_2'(y)=\int_{5/2}^y\frac{-t^2-t+1}{\sqrt{(t+2)(t^2-2)}}\,dt=I_2(5/2)-I_2(y)$$ Note that $I_2$ has no meaning by itself, but only as a part of $I_2'$. The infinite part in $I_2'$ works out to $-\frac23\sqrt{\frac{y^2-2}{y+2}}(1+\sqrt2)\sqrt{3-2\sqrt2}(y+1)$; when combined with the infinite part of $I_1$, their sum tends to zero as $y\to\infty$. Now $$I_0=\frac{\sqrt2}4(I_1(\infty)+I_1(5/2)+I_2(5/2)-I_2(\infty))$$ $$=\frac{\sqrt2}4\cdot\frac{\sqrt{4-2\sqrt2}}3\left( \begin{align} &K(m_1)-(2+\sqrt2)E(m_1)\\ +&F(\varphi_1|_{y=5/2},m_1)-(2+\sqrt2)E(\varphi_1|_{y=5/2},m_1)+\sqrt{\frac{1193\sqrt2+1954}{136}}\\ +&(1+\sqrt2)F(\varphi_2|_{y=5/2},m_2)-(2+\sqrt2)E(\varphi_2|_{y=5/2},m_2) + 7\sqrt{\frac{17\sqrt2+34}{72}}\end{align}\right)$$ $$=\frac{\sqrt{2-\sqrt2}}6(K(m_1)-(2+\sqrt2)E(m_1)+F(\psi_1,m_1)-(2+\sqrt2)E(\psi_1,m_1)+(1+\sqrt2)F(\psi_2,m_2)-(2+\sqrt2)E(\psi_2,m_2))+\sqrt{\frac{2518+300\sqrt2}{1377}}$$ $$\psi_1=\sin^{-1}\sqrt{\frac{5+2\sqrt2}{17}}\qquad\psi_2=\sin^{-1}\frac{\sqrt{2\sqrt2+4}}3$$