Eureka!

$\begin{aligned} \int_0^\infty \frac{\ln x}{x^{\frac23}(1+x)}\,dx & = \; \frac{d}{d\alpha} \int_0^\infty \frac{x^{\alpha-1}}{1+x}\,dx \Big|_{\alpha = \frac13} \; = \; \frac{d}{d\alpha}B(\alpha,1-\alpha)\Big|_{\alpha = \frac13} \\ & = \; \frac{d}{d\alpha}\big(\Gamma(\alpha)\Gamma(1-\alpha)\big)\Big|_{\alpha=\frac13} \; = \; \frac{d}{d\alpha}\big(\pi \,\mathrm{cosec}\pi \alpha\big)\Big|_{\alpha = \frac13} \\ & = \; -\pi^2 \mathrm{cosec}\,\pi\alpha\,\cot\pi\alpha\Big|_{\alpha = \frac13} \; = \; -\tfrac23\pi^2 \end{aligned}$ making the answer $2 + 3= \boxed{5}$ .