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This problem was proposed by Cornel and he showed that

$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)$$

here is my approach.

we know that $\quad\zeta(2)-H_{2n}^{(2)}=\psi^{(1)}(2n+1)=-\displaystyle\int_0^1\frac{x^{2n}\ln x}{1-x}\ dx$

then $$S=-\int_0^1\frac{\ln x}{1-x}\sum_{n=1}^\infty\frac{x^{2n}H_n}{n}\ dx=-\int_0^1\frac{\ln x}{1-x}\left(\frac12\ln^2(1-x^2)+\operatorname{Li}_2(x^2)\right)\ dx$$ and by applying IBP , we get $$S=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx$$ I applied the dilogarithmic identity $\quad\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$

but it was not that helpful. any idea?

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in the body, I reached \begin{align} S&=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1-x^2)}{x(1-x^2)}\ dx\\ &=-2\int_0^1\frac{\operatorname{Li}_2(1-x)\left[\ln(1-x)+\ln(1+x)\right]}{x(1-x)(1+x)}\ dx,\qquad 1-x=y\\ &=-2\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln x+\ln(2-x)\right]}{x(1-x)(2-x)}\ dx\\ &=-2\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{x(1-x)(2-x)}\ dx-2\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x(1-x)(2-x)}\ dx\\ &=-2\left(I_1+I_2\right) \end{align} using the partial fraction decomposition $\quad\displaystyle\frac{1}{x(1-x)(2-x)}=\frac1{1-x}+\frac12\left(\frac1{x}-\frac1{2-x}\right)$, we get \begin{align} I_1&=\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{x}\ dx-\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln x}{2-x}\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\int_0^1\frac{x^n\ln x}{1-x}\ dx+\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\ln x\ dx-\frac12\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln x\operatorname{Li}_2(x)\ dx\\ &=-\sum_{n=1}^\infty\frac{\zeta(2)-H_n^{(2)}}{n^2}-\frac12\sum_{n=1}^\infty\frac1{n^4}-\frac12\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2H_n}{n^3}+\frac{H_n^{(2)}}{n^2}-\frac{2\zeta(2)}{n^2}\right)\\ &=-\zeta^2(2)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}-\frac12\zeta(4)-\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}+\zeta(2)\operatorname{Li}_2\left(\frac12\right) \end{align} plugging the value of the second and third sum proved in here and here respectively, along with the value of the first sum $\frac74\zeta(4)$ and $\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$, we get $$\color{red}{I_1=-\frac1{16}\ln^42-\frac38\ln^22\zeta(2)-\frac{5}{32}\zeta(4)-\frac32\operatorname{Li}_4\left(\frac12\right)}$$ using the partial fraction decomposition for the second integral too, we get, \begin{align} I_2&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{1-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x}\ dx-\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{2-x}\ dx\\ &=A+\frac12B-\frac12C \end{align} To evaluate $A\ $, let $1-x=y$ then apply IBP, we get \begin{align} A&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{1-x}\ dx=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln(1+x)}{x}\ dx\\ &=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}\ dx=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^n\ln x}{1-x}\ dx\\ &=-\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\zeta(2)-H_n^{(2)}\right)\\ &=-\zeta(2)\operatorname{Li}_2(-1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^2} \end{align} substituting the closed form of the sum calculated here along with $\operatorname{Li}_2(-1)=-\frac12\zeta(2)$, we get

$$\color{blue}{A=-\frac16\ln^42+\ln^22\zeta(2)-\frac72\ln2\zeta(3)+\frac{71}{16}\zeta(4)-4\operatorname{Li}_4\left(\frac12\right)}$$ \begin{align} B&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln2+\ln(1-x/2)\right]}{x}\ dx\\ &=\ln2\operatorname{Li}_3(1)-\sum_{n=1}^\infty\frac{1}{n2^n}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\ln2\zeta(3)-\sum_{n=1}^\infty\frac{1}{n2^n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\ln2\zeta(3)-\zeta(2)\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n}{n^32^n} \end{align} plugging the well known result of the sum and $\operatorname{Li}_2\left(\frac12\right)$, we get $$\color{blue}{B=\frac1{24}\ln^42+\frac12\ln^22\zeta(2)+\frac78\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)}$$ \begin{align} C&=\int_0^1\frac{\operatorname{Li}_2(x)\ln(2-x)}{2-x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\left[\ln2+\ln(1-x/2)\right]}{2-x}\ dx\\ &=\ln2\int_0^1\frac{\operatorname{Li}_2(x)}{2-x}\ dx+\frac12\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x/2)}{1-x/2}\ dx\\ &=\ln2\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx-\sum_{n=1}^\infty\left(\frac{H_n}{2^n}-\frac{1}{2^n}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\ln2\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\sum_{n=1}^\infty\left(\frac{H_n}{2^n}-\frac{1}{n2^n}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{n^32^n}-\ln2\sum_{n=1}^\infty\frac{H_n}{n^22^n}-\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n2^n}+\ln^22\zeta(2)+\zeta(2)\operatorname{Li}_2\left(\frac12\right) \end{align} substituting the value of the first sum ( can be found here ) and the values of the other well known sums, we get $$\color{blue}{C=-\frac1{12}\ln^42+\frac54\ln^22\zeta(2)-\frac{21}8\ln2\zeta(3)+\frac{35}{16}\zeta(4)-2\operatorname{Li}_4\left(\frac12\right)}$$ by grouping $A$, $B$ and $C$, we get $$\color{red}{I_2=-\frac5{48}\ln^42+\frac58\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{89}{32}\zeta(4)-\frac52\operatorname{Li}_4\left(\frac12\right)}$$ Finally \begin{align} S&=-2\left(I_1+I_2\right)\\ &=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+8\operatorname{Li}_4\left(\frac12\right) \end{align}

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