Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$

The substitution $t = (1 + x^3)^{-1}$ yields $$ I = \frac{1}{3} \int \limits_0^1 - \ln(t) \left(\frac{t}{1-t}\right)^{2/3} \, \mathrm{d} t = f'\left(\frac{2}{3}\right) \, , $$ where $$ f(\alpha) \equiv - \frac{1}{3} \int \limits_0^1 \frac{t^\alpha}{(1-t)^{2/3}} \, \mathrm{d} t = - \frac{1}{3} \operatorname{B} \left(\frac{1}{3}, \alpha +1 \right) = - \frac{\operatorname{\Gamma} \left(\frac{1}{3}\right)}{3} \frac{\operatorname{\Gamma}(\alpha + 1)}{\operatorname{\Gamma}\left(\alpha + \frac{4}{3} \right)} ~~~ , \, \alpha > -1 \, .$$ The differentiation under the integral sign can be justified using the dominated convergence theorem.

In terms of the digamma function $\psi$ we now have $$ I = f'\left(\frac{2}{3}\right) = \frac{2}{9} \operatorname{\Gamma} \left(\frac{1}{3}\right) \operatorname{\Gamma} \left(\frac{2}{3}\right) \left[\operatorname{\psi} (2) - \operatorname{\psi} \left(\frac{5}{3}\right)\right] \, . $$ We can use the reflection formulas for $\Gamma$ and $\psi$ and the recurrence relation $\operatorname{\psi}(x + 1) = \operatorname{\psi}(x) + \frac{1}{x}$ to find $$ I = \frac{2}{9} \frac{\pi}{\sin\left(\frac{\pi}{3}\right)} \left[\operatorname{\psi}(1) - \operatorname{\psi} \left(\frac{1}{3}\right) - \pi \cot\left(\frac{\pi}{3}\right) - \frac{1}{2} \right] \, .$$ The special values $\operatorname{\psi}(1) = - \gamma$ and $\operatorname{\psi} \left(\frac{1}{3}\right) = - \frac{\pi}{2 \sqrt{3}} - \frac{3}{2} \ln (3) - \gamma$ then lead to the final result $$ I = \frac{2 \pi}{27} [\sqrt{3} (3 \ln(3) - 1) - \pi] \, .$$

In this solution we will not use the beta function or the digamma function.

Put \begin{equation*} f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x. \end{equation*} We want to calculate $f(1)$. However, \begin{equation*} f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1} \end{equation*} and \begin{equation*} f'(s)=\int_{0}^{\infty}\dfrac{3s^2x^3}{(1+s^3x^3)(1+x^3)^2}\, \mathrm{d}x =[x^3=z] = \int_{0}^{\infty}\dfrac{s^2z^{1/3}}{(1+s^3z)(1+z)^2}\, \mathrm{d}z . \end{equation*} For $0<s<1$ we use keyhole contour integration and get \begin{equation*} f'(s) = \dfrac{2\pi\sqrt{3}}{9}\dfrac{2s+1}{(s^2+s+1)^2}\cdot s^2 . \end{equation*} Integration by parts in (1) will finally give us \begin{equation*} f(1) = \dfrac{2\pi}{27}\left(\sqrt{3}(3\ln 3 -1)-\pi\right). \end{equation*}