<s>[INST] A really interesting logarithm integral I need to prove that 

$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x  = 
 \frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$

My approach 
We know that 
$$\int^1_0 \frac{\log x}{x^2-2\cos(\theta)x+1}\mathrm{d}x =- \frac{\mathrm{cl}_2(\theta)}{\sin(\theta)}$$
Let $\theta=\pi/3$
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x =- \frac{2}{\sqrt{3}}\mathrm{cl}_2(\pi/3)$$
By definition 
\begin{align}
\mathrm{cl}_2(\pi/3) &= -\int^{\pi/3}_0 \log(2\sin \frac{t}{2})\mathrm{d}t\\
& = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\sin(\pi x))\,\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\pi)-\log\Gamma(x)-\log\Gamma(1-x)\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+\log\Gamma(1-x)\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+2\pi\int^{1}_{5/6}\log\Gamma(x)\mathrm{d}x
\end{align}
Now use the loggamma integral  
$$\int_{0}^{z} \log \Gamma(x)  \,  \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$
Implies the two integrals
$$\int_{0}^{1/6} \log \Gamma(x)  \,  \mathrm dx = \frac{1}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,1/6)$$
$$\int_{5/6}^{1} \log \Gamma(x)  \,  \mathrm dx =\frac{1}{2}\log(2\pi)-( \frac{5}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,5/6))$$
Hence we have 
$$\mathrm{cl}_2(\pi/3) = -\frac{\pi}{3}\log(2\pi)+2\pi (\frac{1}{6}\log(2\pi)+\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6)) \\=2\pi (\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6)) $$
This implies that 
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x  = \frac{-4\pi}{\sqrt{3}}\left(\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6) \right) = \frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$
Maybe using 
$$ \zeta(s,p/q)=2\Gamma(1-s)(2\pi q)^{s-1}\sum_{n=1}^q \sin \left[\frac{\pi s}{2}+ \frac{2 \pi n p}{q}\right] \zeta(1-s,n/q) $$
Question 


*

*My approach seems to be so indirect I am really interested in seeing more directed approaches that don't use loggamma integral.

*I am not sure about the last step (relating the derivative of the Hurwitz zeta function to the trigamma. 
 [/INST] Hint. One may write
$$
\int^1_0 \frac{\log x}{x^2-x+1}dx=\int^1_0 \frac{(1+x)\log x}{1+x^3}dx=\left.\partial_s\int^1_0 \frac{(1+x)\:x^s}{1+x^3}dx\right|_{s=0}
$$ then after a change of variable one may use
$$
\int^1_0 \frac{t^s}{1+t}dt=\frac12\psi\left(\frac{s}2+1\right)-\frac12\psi\left(\frac{s}2+\frac12\right), \quad s>-1.
$$
</s>