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After reading some meta posts, I've decided to post this question on MathOverflow since I didn't receive any comments or answers on MSE

Certainly, I apologize for any oversight. Here's a more refined version:

Integral to Evaluate:

$$\int_0^1 \dfrac{\operatorname{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\dfrac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx=\dfrac{\pi^4}{1944}.$$

Approach:

Utilize the substitution $t=x/4$, integration by parts, and the identity $\sum_{n=1}^{\infty}H_{n}^{(2)}x^{n}=\frac{Li_{2}(x)}{1-x}$. Consider breaking it up using:

$$H_{n+1}^{(2)}-\frac{1}{(n+1)^{2}}=H_{n}^{(2)},$$

and/or

$$\sum_{n=1}^{\infty}\frac{x^{2n}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4(\sin^{-1}(\frac{x}{2}))^{2}}{x^{2}}.$$

This is related to the identity:

$$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=1}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}} \quad |z|<1.$$

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    $\begingroup$ Why downvotes? Can someone explain $\endgroup$ Commented Feb 24, 2024 at 13:42
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    $\begingroup$ You can improve your post by adding a high-level tag and delete personal comments (life stories) like "I am going to have a look", "It's been a while", "Unless, someone wants to jump in", "I get it now". This is not a chat site. Ask a question concisely, and provide some context. Regarding high-level tags: meta.mathoverflow.net/questions/1075/… $\endgroup$ Commented Feb 24, 2024 at 13:50
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    $\begingroup$ @GHfromMO sorry sir, I'm a new user on MathOverflow, and I edited the question, but I'm not sure if it was a good edit. $\endgroup$ Commented Feb 24, 2024 at 13:58
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    $\begingroup$ Why do you think your integral equals $\pi^4/1944$? $\endgroup$ Commented Feb 24, 2024 at 15:04
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    $\begingroup$ As the OP clearly noticed, his formula is equivalent to the identity $$\sum_{n\ge1}H_n^{(2)}/((n+1)(2n+1)\binom{2n}{n})=\pi^4/972$$ which is probably known and easier to prove. $\endgroup$ Commented Feb 24, 2024 at 17:00

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As Henri Cohen remarked, the identity to prove is equivalent to $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{\pi^4}{972}.\tag{1}$$ In turn, this follows readily from the OP's last display (which is a known identity, see e.g. here): $$\left( \sin^{-1}(z)\right)^4=\frac{3}{2}\sum_{k=2}^\infty\frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2 \binom{2k}{k}}, \qquad |z|<1.\tag{2}$$ Let us see how $(2)$ implies $(1)$. We write $k=n+1$ in $(2)$, and observe that $$(n+1)^2\binom{2n+2}{n+1}=2(n+1)(2n+1)\binom{2n}{n}.$$ Therefore, $$\left( \sin^{-1}(z)\right)^4=\frac{3}{4}\sum_{n=1}^\infty\frac{H_n^{(2)}(2z)^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}}, \qquad |z|<1.$$ For $z=1/2$ this gives $(1)$: $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\binom{2n}{n}}=\frac{4}{3}\left( \sin^{-1}\Bigl(\frac{1}{2}\Bigr)\right)^4=\frac{4}{3}\Bigl(\frac{\pi}{6}\Bigr)^4=\frac{\pi^4}{972}.$$

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