The question here requires the evaluation of the following summation:
$$\sum_{n=1}^\infty\frac 1{(2n+1)(2n+2)}\sum_{r=1}^n \frac 1r\tag{1}$$
The answers provided (and also wolframalpha) show that the solution is $$\frac {\pi^2}{12}-(\ln 2)^2\tag{2}$$
From the standard expansion of $\ln (1+x)$ and result of the Basel problem, $(2)$ can be written as $$\sum_{n=1}^\infty \frac 1{2n^2}-\left(\sum_{n=1}^\infty \frac {(-1)^{n+1}}n\right)^2\tag{3}$$ and from answers to my other question on the Cauchy Product here, the second term above can be written as $$\left(\sum_{n=1}^\infty \frac {(-1)^{n+1}}n\right)^2=\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{j+1}}j\cdot \frac {(-1)^{n+2-j}}{n+1-j} =\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}$$ Hence $(3)$ can be written as $$\sum_{n=1}^\infty \frac 1{2n^2}-\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}=\sum_{n=1}^\infty \left(\frac 1{2n^2}-\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}\right)\tag{4}$$
Question
Is it possible to derive $(3)$ or $(4)$ directly from $(1)$ without first working out the closed-form result given in $(2)$?
