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Are we aware of any nice way of calculating these $2$ integrals?

$$i) \space \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{x^2-x+1} \, dx$$

$$ii)\space \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{\left(x^2-x+1\right)^2} \, dx$$

If considering in the first integral that $$\frac{1}{{x^2-x+1}}=\frac{1}{i \sqrt{3}}\left(\frac{1}{ \left(x-\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)\right)}-\frac{1}{ \left(x-\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\right)}\right)$$ then it seems Mathematica can handle the resulting integrals.

It's also easy to note that $$\frac{1}{i \sqrt{3}}\int_0^1 \frac{(1-2x)\log \left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right) \log \left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x(1-x)} \, dx=0$$ and then for the remaining part Mathematica yields some shorter result in polylogarithms. So, we have that $$ \int_0^1 \frac{\text{Li}_2\left(x-x^2\right)}{x^2-x+1} \, dx$$ $$=\frac{i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx-\frac{i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx$$ $$=\frac{2i}{\sqrt{3}}\int_0^1 \frac{ (1-2 x) \log ^2\left(x-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{ x-x^2} \, dx$$ where the last equality is simply got by using the variable change $x\mapsto 1-x$.

EDIT: Using the idea of dilogarithm reflection formula, we get that

$$\int_0^1 \frac{\text{Li}_2\left(x^2-x+1\right)}{x^2-x+1} \, dx=\frac{2}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(\frac{3 \sec ^2(x)}{4}\right) \, dx$$ $$=\frac{4}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(-\frac{\sqrt{3}}{2} \sec (x)\right) \, dx+\frac{4}{\sqrt{3}} \int_{-\pi/6}^{\pi/6} \text{Li}_2\left(\frac{\sqrt{3}}{2} \sec (x)\right) \, dx$$ but not sure yet if it might lead to an elegant solution.

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A first step. Let: $$ I_k = \int_{0}^{1}\frac{(x-x^2)^k}{1-x+x^2}\,dx.$$ We have $I_0=\frac{2\pi}{3\sqrt{3}}, I_1=-1+\frac{2\pi}{3\sqrt{3}}$ and: $$ I_{n+1} = \int_{0}^{1}\frac{(x-x^2)(x-x^2)^n}{1-x+x^2}\,dx = I_n-\int_{0}^{1}x^n(1-x)^n\,dx = I_n - \frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$ hence: $$ I_{n}=\frac{2\pi}{3\sqrt{3}}-\sum_{k=0}^{n-1}\frac{1}{(2k+1)\binom{2k}{k}}\tag{1}$$ and: $$\begin{eqnarray*} \mathcal{J}_1=\int_{0}^{1}\frac{\text{Li}_2(x-x^2)}{1-x+x^2}\,dx &=& \frac{\pi^3}{9\sqrt{3}}-\sum_{n\geq 1}\frac{1}{n^2}\sum_{k=1}^{n}\frac{\Gamma(k)^2}{\Gamma(2k)}\\&=&\frac{\pi^3}{9\sqrt{3}}-\sum_{k\geq 1}\frac{\Gamma(k)^2}{\Gamma(2k)}\left(\zeta(2)-H^{(2)}_{k-1}\right)\\&=&\sum_{k\geq 1}\frac{\Gamma(k)^2}{\Gamma(2k)}H_{k-1}^{(2)}.\tag{2}\end{eqnarray*}$$ Maybe it is useful to recall that $$\frac{\Gamma(k)^2}{\Gamma(2k)} = [z^{2k-1}]\frac{2\arcsin(z/2)}{\sqrt{1-z^2/4}},\qquad H_{k-1}^{(2)}=[z^k]\frac{z\cdot\text{Li}_2(z)}{1-z}.$$ According to (a carefully driven) Mathematica, we simply have:

$$ \mathcal{J}_1 = \frac{\pi^3}{81\sqrt{3}},\tag{3}$$

quite astonishing. I bet that the reflection formula $$ \text{Li}_2(x)+\text{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x)$$ and the other dilogarithm functional identities are behind this conspiracy, just cannot grasp how, at the moment.

As a matter of fact, that is not too difficult. If we complete the OP's manipulations with integration by parts, we are left to computing the two "conjugated" integrals: $$ \int_{0}^{1}\frac{\log(x)\log(x-\omega)}{x-\omega}\,dx,\qquad \int_{0}^{1}\frac{\log(1-x)\log(x-\omega)}{x-\omega}\,dx$$ where the first one, for instance, is just: $$ \frac{i}{81}\left(2\pi^3-27\pi\,\text{Li}_2(\overline{\omega})+54i\,\zeta(3) \right)$$ by the residue theorem. As an alternative, we can deal with $(2)$ through the Flajolet-Salvy technique. In a similar fashion,

$$\begin{eqnarray*} \mathcal{J}_2 &=& \int_{0}^{1}\frac{\text{Li}_2(x-x^2)}{(1-x+x^2)^2}\,dx\\&=&3+\frac{\pi^2}{9}-\frac{16}{3}\log 2-\frac{32}{3\sqrt{3}}\int_{0}^{1}\frac{x}{1-x^2}\arctan\left(\frac{x}{\sqrt{3}}\right)\log\left(\frac{1+x}{2}\right)\,dx.\end{eqnarray*} $$

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