Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:

Prove that  $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $

where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma function.
I've tried reflection formula and some Feynmann method on the integrals in the wikipedia but nothing works.
(I've copied this from AoPS because there's no response there)

Let $(A_n)$ and $(B_n)$ by
$$ A_n = \frac{1^1 2^2 \cdots n^n}{n^{n^2/2+n/2+1/12} e^{-n^2/4}}
\quad \text{and} \quad
B_n = \frac{n!}{e^{-n}n^{n+1/2}}. $$
In view of the Wikipedia page, we know that $A_n \to A$ and $B_n \to \sqrt{2\pi}$. We can also equivalently state these definitions as follows:
\begin{align*}
\sum_{k=1}^{n} k \log k
&= \log A_n + \left( \frac{n^2}{2} + \frac{n}{2} + \frac{1}{12} \right) \log n - \frac{n^2}{4}, \\
\sum_{k=1}^{n} \log k
&= \log B_n + \left(n + \frac{1}{2}\right) \log n - n.
\end{align*}
Now using the partial fractional decomposition of $\psi(x)$, we find that
\begin{align*}
\int_{0}^{1} x^2 \psi(x) \, dx
&= \lim_{n\to\infty} \int_{0}^{1} x^2 \left( -\gamma + H_{n+1} - \frac{1}{x} - \sum_{k=1}^{n} \frac{1}{k+x} \right) \, dx \\
&= \lim_{n\to\infty} \left( \frac{H_{n+1}-\gamma}{3} + \frac{n^2}{2} - \frac{1}{2} + \sum_{k=1}^{n} k^2 (\log k - \log(k+1)) \right) \\
&= \lim_{n\to\infty} \left( \frac{1}{3}\log n + \frac{n^2}{2} - \frac{1}{2} + \sum_{k=1}^{n} (2k-1)\log k - n^2 \log(n+1) \right) \\
&= \lim_{n\to\infty} \left(2\log A_n - \log B_n - n^2 \log\left(1+\frac{1}{n}\right) + n - \frac{1}{2} \right) \\
&= 2\log A - \log\sqrt{2\pi}.
\end{align*}
