How many integrals ?

One of the approach which I wish to share. $\int_{0}^1\frac{(x-\sqrt[3]{x})\ln(x+\sqrt[3]{x})}{x+\sqrt[3]{x}}dx \overbrace{=}^{x=u^3}3\int_0^1\frac{u(u^3-u)\ln(u^3+u)}{u^2+1}du$ the latter integral we write as $3^{-1}I=\int_0^1\left(\frac{u^4\ln(u^3+u)}{u^2+1}-\frac{u^2\ln(u^3+u)}{u^2+1}\right)$ let $u=\tan y$ then the integral becomes $\int_0^{\frac{\pi}{4}}\tan^4y\ln(\tan y \sec^2y)-\tan^2y\ln(\tan y \sec^2y)dy$ $\underbrace{\int_0^{\frac{\pi}{4}}\left(\tan^4y-\tan^2y\right)\ln(\tan y)dy}_{I_1}+\underbrace {\int_0^{\frac{\pi}{4}}(\tan^4y-\tan^2 y)\ln(\sec^2 y)dy}_{I_2}$ By using the reduction formula of $I(n)=\displaystyle \int\tan^n ydy$ for $n=4$ we deduce that $I(4)=\int \tan^4 y=y+\frac{1}{3} \sec^2y \tan y-\frac{4}{3}+C$ and $n=2$ is trivial to see $\displaystyle I(2)=\int\tan^2 y=\tan y-y+C$ Now we calculate $I_1,I_2$ using integration by parts $I_1=\underbrace{\ln(\tan y)\int_0^{\frac{\pi}{4}}(\tan^4y-\tan^2 y)dy}_{0}-\int_0^{\frac{\pi}{4}}\frac{\sec^2y}{\tan y}(I(4)-I(2))dy$ Further $I_1=\int_0^{\frac{\pi}{4}}\frac{\sec^2 y}{\tan y}\left(2y+\frac{\tan y}{3}\left(\sec^2y-7\right)\right)dy$ $=\underbrace {\int_0^{\frac{\pi}{4}} 2\left(y\tan y +y\cot y\right)dy}_{I_3}+\underbrace {\frac{1}{3}\int_{0}^{\frac{\pi}{4}} \sec^2y(\sec^2 y-7)dy}_{I_4}$ Here $I_3=\overbrace{y\int_0^{\frac{\pi}{4}}(\tan y+\cot y)}^{0}-\int_{0}^{\frac{\pi}{4}}\left(\int(\tan y +\cot y)dy\right)dy$ $=-\int_0^{\frac{\pi}{4}}\ln(\tan y)dy \overbrace{=}^{\tan y=t}- \int_0^1\frac{\ln t}{1+t^2}dt =-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n-1)^2}=G\Rightarrow I_3=-2G$ Further we evaluate $3I_4=\int_0^{\frac{\pi}{4}} \sec^2y(\sec^2y-7)dy=\int_0^{\frac{\pi}{4}}(\sec^2y\tan^2y-6\tan^2y)$ $=\left[\frac{\tan^3y}{3}-6\tan y\right]_0^{\frac{\pi}{4}}= -\frac{17}{3}\Rightarrow I_4=\frac{17}{9}$ and hence $I_1=\int_0^{\frac{\pi}{4}}\left(\tan^4y-\tan^2y\right)\ln(\tan y)dy=\frac{17}{9}-2G$ Now we evaluate $I_2$ $\int_0^{\frac{\pi}{4}}(\tan^4y-\tan^2 y)\ln(\sec^2 y)dy=-2\int_0^{\frac{\pi}{4}}(\tan^4y-\tan^2 y)\ln(\cos y)dy$ $\overbrace{=}^{IBP}-2\ln(\cos y)\int_0^{\frac{\pi}{4}}(\tan^4-\tan^2 y)dy+2\int_0^{\frac{\pi}{4}}\left( (-\tan y)\int(\tan^4 y-\tan^2 y)dy\right)dy$ $=-\frac{5}{3}\ln2+\frac{\pi}{2}\ln2-2\int_0^{\frac{\pi}{4}}\tan y\left(2y+\frac{1}{3}\sec^2 y\tan y-\frac{7}{3}\tan y\right)dy$ Further $\int_0^{\frac{\pi}{4}}\left(-4y\tan y-\frac{2}{3}\sec^2y\tan^2y+\frac{14}{3}\tan^2y\right)dy=-4\int_0^{\frac{\pi}{4}}y\tan y +\frac{40}{9}-\frac{7\pi}{6}$

$=-4y\int_0^{\frac{\pi}{4}}\tan y +4\int_0^{\frac{\pi}{4}}\left(\int\tan y dy\right)dy=-\frac{\pi}{2} \ln 2-4\int_0^{\frac{\pi}{4}}\ln(\cos y)dy$ Let $\displaystyle J_1 =\int_0^{\frac{\pi}{4}}\ln(\cos y)dy \; ,J_2=\int_0^{\frac{\pi}{4}}\ln(\sin y) dy$ Note that $J_1+J_2$ $=\int_0^{\frac{\pi}{4}}\ln\left(\frac{1}{2}\sin2 y\right)dy\overbrace{=}^{2y=t}\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln(\sin t)dt -\frac{\pi}{4}\ln 2 =-\frac{\pi}{4}\ln 2 -\frac{\pi}{4}\ln 2 =-\frac{\pi}{2}\ln 2$ Also $J_2-J_1=\int_0^{\frac{\pi}{4}}\ln(\tan y)dy\overbrace{=}^{ I_1} -G$ So $\displaystyle=-\frac{\pi}{2}\ln 2-4\int_0^{\frac{\pi}{4}}\ln (\cos y)dy= -\frac{\pi}{2}\ln 2 -4\left(\frac{G}{2}-\frac{\pi}{2}\ln2\right)=\frac{\pi}{2}\ln 2 -2G$ Making the finally answer $3(I_1+I_2)$ $\displaystyle= 3\left(\frac{40+17}{9}-4G+\pi\ln 2-\frac{5}{3}\ln 2 -\frac{7\pi}{3}\right)=19-12G+3\pi\ln2 -5\ln(2) -\frac{7\pi}{2}\approx 0.07986095042\cdots$

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The solution I had is quite lengthy and to speed up the calculation and reduce the work I even tried up with generating function for harmonic numbers which seems to be tedious again with calculations. So im looking for any short solution/s. I will be glad to know if anyone wish to share their approach.