Evaluate $\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$ I would like to evaluate this integral.
$$\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$$
$1+x+x^2=(x+\frac{1}{2})^2+1-\frac{1}{4}=(x+\frac{1}{2})^2-\frac{3}{4}$
$$\int_{0}^{1}\frac{x(1-x)^2}{(x+\frac{1}{2})^2-\frac{1}{4}}\frac{\mathrm dx}{\ln x}$$
$$-\int_{0}^{\infty}\frac{e^{2y}(1-e^y)^2}{(e^y+\frac{1}{2})^2-\frac{1}{4}}\frac{\mathrm dy}{y}$$
I did a useless substitution, I was hoping to get a simpler integral.
 A: 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*}
