I'm trying to prove $$I=\iint_{[0,1]^2}\frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\frac83(\sqrt2-1)$$ in an elementary way.
What I've tried
I'm not satisfied with my following method.
Integrating w.r.t. $x$:
$$I=\int_0^1\left[-\frac{20}3 \left(y^{3/2}-\sqrt{y^3+1}\right)-y^{-3/2}\, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{y^3}\right)\right]dy$$
Integrating w.r.t. $y$:
$$I=\int_0^1\left(10 \sqrt x-x^{-3/2}\right) \, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{x^3}\right)dx$$
Although I know how to handle some integrals of kind $\int_0^1x^\alpha(1-x)^\beta {}_pF_q(x)dx$ and how to cancel some "adjacent" hypergeometric function values, but concerning the answer and the integrand is so simple, I strongly believe that there are simple substitutions , IBPs, Feymann's tricks or something similar that can directly solve the integral. Any alternative solution will be appreciated.
Best Answer
$$ I=\int_0^1\int_0^1 \frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\int_0^1\int_0^1 \frac{10y^2-1}{\sqrt{x^3+y^3}}dxdy$$ $$\Rightarrow I=5\int_0^1\int_0^1 \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-\int_0^1\int_0^1 \frac{1}{\sqrt{x^3+y^3}}dxdy$$ $$=10\int_0^1 \int_0^y \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-2\int_0^1\int_0^y \frac{1}{\sqrt{x^3+y^3}}dxdy$$ $$\overset{x=ty}=10\int_0^1 y \sqrt y\int_0^1\frac{t^2+1}{\sqrt{t^3+1}}dtdy-2\int_0^1\frac{1}{\sqrt y}\int_0^1 \frac{1}{\sqrt{t^3+1}}dtdy$$ $$=4\int_0^1 \frac{t^2+1}{\sqrt{t^3+1}}dt-4\int_0^1 \frac{1}{\sqrt{t^3+1}}dt=4\int_0^1 \frac{t^2}{\sqrt{t^3+1}}dt$$ $$=\frac83\sqrt{t^3+1}\bigg|_0^1=\frac83\left(\sqrt 2-1\right)$$