Evaluate $\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$

Consider the function

$$ I(s) = \int_{0}^{1} \frac{x^{s+1} (1 - x)^2}{1+x+x^2}\,\frac{\mathrm{d}x}{\log x}. $$

Then

\begin{align*} I'(s) &= \int_{0}^{1} \frac{x^{s+1} (1 - x)^3}{1-x^3}\,\mathrm{d}x \\ &= \frac{1}{3}\int_{0}^{1} \frac{u^{(s-1)/3} (1 - u^{1/3})^3}{1-u}\,\mathrm{d}u \tag{$x=u^{1/3}$} \\ &= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \int_{0}^{1} \frac{1-u^{(s+k-1)/3}}{1-u}\,\mathrm{d}u \\ &= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \psi((s+k+2)/3), \end{align*}

where we utilized the identity $\sum_{k=0}^{3}\binom{3}{k}(-1)^k = 0$. Also, $\psi$ is the digamma function and we utilized the identity that $\int_{0}^{1} \frac{1-u^z}{1-u} \, \mathrm{d}z = \gamma + \psi(z+1)$ in the final step. Together with $I(\infty) = 0$, we get

\begin{align*} I(0) &= -\lim_{R\to\infty}\int_{0}^{R} I'(s) \, \mathrm{d}s \\ &= \lim_{R\to\infty}\sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \left[ \log\Gamma((R+k+2)/3) - \log\Gamma((k+2)/3) \right] \\ &= - \sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \log\Gamma((k+2)/3) \\ &= \log(18) - 3\log\Gamma(1/3). \end{align*}