Methods to solve $\int _0^{\infty }\frac{x^{\frac{4}{5}}-x^{\frac{2}{3}}}{\ln \left(x\right)\left(x^2+1\right)}\:dx$

Consider $$J(a)=\int_0^\infty\frac{x^a-1}{(x^2+1)\ln x}dx$$ where $a\in\Bbb C$ with $|\Re a|<1$. Then $$J'(a)=\int_0^\infty\frac{x^a}{x^2+1}dx.$$ Using the keyhole contour you can see that $$J'(a)=\frac{2\pi i}{1-e^{2\pi a}}\left(\frac{e^{\frac{\pi a}{2}}}{2i}+\frac{e^{\frac{3\pi a}{2}}}{-2i}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}.$$

Alternatively with $y=x^2$ and $u=\frac{y}{y+1}$, note that \begin{align}J'(a)&=\frac{1}{2}\int_0^\infty \frac{y^{\frac{a+1}{2}-1}}{y+1}dy=\frac12\int_0^1 u^{\frac{a+1}{2}-1}(1-u)^{\left(1-\frac{a+1}{2}\right)-1}du\\&=\frac12\mathrm{B}\left(\frac{a+1}{2},1-\frac{a+1}{2}\right)=\frac12\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(1-\frac{a+1}{2}\right).\end{align} From Euler's reflection formula, $J'(a)=\frac{\pi}{2}\operatorname{cosec}\left(\pi\frac{a+1}{2}\right)=\frac{\pi}{2}\sec\frac{\pi a}{2}$.

Therefore $$J(a)=\int_0^a \frac{\pi}{2}\sec \frac{\pi t}{2} dt=\int_0^{\frac{\pi a}{2}} \sec\theta d\theta=\ln\left(\tan\frac{\pi a}{2}+\sec\frac{\pi a}{2}\right).$$ Therefore $$I=J\left(\frac45\right)-J\left(\frac23\right)=\ln\frac{\tan\frac{2\pi}{5}+\sec \frac{2\pi}{5}}{\tan\frac{\pi}{3}+\sec{\frac{\pi}{3}}}=\ln\frac{\sqrt{5+2\sqrt{5}}+1+\sqrt{5}}{\sqrt3 +2}\approx 0.525772.$$