Trigamma identity $\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$ Regarding this integral: Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ The following conjecture comes: $$\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$$ Where $G$ is Catalan's constant. How can we show this? Wolfram-alpha agrees on this: https://www.wolframalpha.com/input/?i=trigamma(11%2F12)-trigamma(5%2F12)%3D4sqrt3pi%5E2-80Catalan

\begin{align} 
\psi_1(\tfrac{11}{12})
-\psi_1(\tfrac{5}{12})
&=4\sqrt 3 \pi^2-80G
\tag{1}\label{1}
.
\end{align}  
The constant $G$ - Catalan's constant 
has known to appear in relations 
\begin{align} 
\psi_1(\tfrac14)&=\pi^2+8G
,\\
\psi_1(\tfrac34)&=\pi^2-8G
,
\end{align}
so we can try to start from $\psi_1(\tfrac14)$.
Applying triplication identity 
\begin{align}
9\psi_1(3x) &= 
\psi_1(x)
+\psi_1(x+\tfrac13)
+\psi_1(x+\tfrac23)
\tag{2}\label{2}
\end{align}
to $\psi_1(\tfrac14)=\psi_1(3\cdot\tfrac1{12})$, 
we get
\begin{align}
9\psi_1(\tfrac14) 
&= 
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
+\psi_1(\tfrac34)
,\\
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
&=9\psi_1(\tfrac14)-\psi_1(\tfrac34)
\\
&=10\psi_1(\tfrac14)-(\psi_1(\tfrac14)+\psi_1(\tfrac34))
\\
&=10\psi_1(\tfrac14)-2\pi^2
\\
&=10(\pi^2+8G)-2\pi^2
\\
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
&=8\pi^2+80G
,
\end{align}
and we are almost there.
Now, applying the identity
\begin{align}
\psi_1(1-x)&=
\frac{\pi^2}{\sin(\pi x)^2}
-\psi_1(x)
\end{align}
to $\psi_1(\tfrac1{12})=\psi_1(1-\tfrac{11}{12})$,
we get
\begin{align}
\psi_1(\tfrac1{12}) 
&=\frac{\pi^2}{\sin(\tfrac\pi{12})^2}
-\psi_1(\tfrac{11}{12})
,\\
&=8\pi^2+4\sqrt3\pi^2
-\psi_1(\tfrac{11}{12})
,
\end{align}
and \eqref{1} follows.
