Show $\int_0^\sqrt[3]2(u^3f(u)+u^7/f(u))du=\ln(1+x)-(\sqrt x-\sqrt2)^2+\frac43$ where $f(u)=\sqrt[3]{27+u^6+\sqrt{27(27+2u^6)}}$ and $x^3-2x^2-4x-2=0$

Question

Inspired by this question, I've come up with an exercise that I hope you will find amusing:

---

Show that

$$\begin{eqnarray*} &&\int_{0}^{\sqrt[3]{2}}\left[u^3\sqrt[3]{27+u^6+\sqrt{27}\sqrt{27+2u^6}}+\frac{u^7}{\sqrt[3]{27+u^6+\sqrt{27}\sqrt{27+2u^6}}}\right]\,du\\&=&\ln(1+\beta)-\left(\sqrt{\beta}-\sqrt{2}\right)^2+\frac{4}{3} \end{eqnarray*}$$

where $\beta$ is the only real solution of $p(x)=x^3-2x^2-4x-2=0$.

The numerical value is approximately $2.6304$.

---

I am leaving in a spoiler how I actually created this monster.

1. $f(x)=\frac{x}{x^2+1}$ is an algebraic function with a small derivative

2. if we rotate the graph of $f(x)$ by $45^\circ$ around the origin we get the graph of an algebraic function $g(x)$ related to the roots of a cubic
   
3. up to some substitution the given integral equals $\int_{0}^{1}g(x),dx$
   
4. which is also the area enclosed by the graph of $f(x)$ and the lines $y=-x,y=x-\sqrt{2}$. Much easier.

• Are you able to prove the statement in a different way, for instance via Lagrange inversion formula?
• Do you know if a similar trick has been used to produce actually interesting identities?