How to prove that $\int_{0}^{1}\ln{(x/(1-x))}\ln{(1+x-x^2)}\frac{dx}{x}=-\frac{2}{5}\zeta{(3)}$

Since $x\left(1-x\right)<1$ if $x\in\left(0,1\right) $ we have $$\begin{align} \int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx & =\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}\left(1-x\right)^{k}dx \\ & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\log\left(1-x\right)x^{k-1}\left(1-x\right)^{k}dx. \end{align} $$ Now by definition of Beta function we have $$ \int_{0}^{1}x^{a}\left(1-x\right)^{k}dx=B\left(a+1,k+1\right) $$ and so $$\begin{align} \int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(x\right)dx = &\frac{\partial}{\partial a}\left(B\left(a+1,k+1\right)\right)_{a=k-1} \\ = & B\left(k,k+1\right)\left(\psi\left(k\right)-\psi\left(2k+1\right)\right) \end{align} $$ and in a similar way we have $$\begin{align} \int_{0}^{1}x^{k-1}\left(1-x\right)^{k}\log\left(1-x\right)dx= & \frac{\partial}{\partial b}\left(B\left(k,b\right)\right)_{b=k+1} \\ = & B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(2k+1\right)\right) \end{align} $$ then $$\begin{align} \int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(-x^{2}+x+1\right)}{x}dx= & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k}B\left(k,k+1\right)\left(\psi\left(k+1\right)-\psi\left(k\right)\right) \\ = & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{2}}B\left(k,k+1\right) \\ = & -\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}} \end{align} $$ and the last series has a well known closed form $$\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k^{3}\dbinom{2k}{k}}=\frac{2}{5}\zeta\left(3\right).\tag{1}$$ For the other integral note that $$I=\int_{0}^{\infty}\frac{\log\left(y\right)\log\left(1+3y+y^{2}\right)}{y\left(y+1\right)}dy\overset{y=\frac{x}{1-x}}{=}\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(\frac{1+x-x^{2}}{\left(1-x\right)^{2}}\right)}{x}dx $$ $$=\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1+x-x^{2}\right)}{x}dx-2\int_{0}^{1}\frac{\log\left(\frac{x}{1-x}\right)\log\left(1-x\right)}{x}dx $$ and it is sufficient to observe that $$-2\int_{0}^{1}\frac{\log\left(x\right)\log\left(1-x\right)}{x}dx=2\sum_{k\geq1}\frac{1}{k}\int_{0}^{1}\log\left(x\right)x^{k-1}dx $$ $$=-2\sum_{k\geq1}\frac{1}{k^{3}}=-2\zeta\left(3\right) $$ and $$ 2\int_{0}^{1}\frac{\log^{2}\left(1-x\right)}{x}dx=2\int_{0}^{1}\frac{\log^{2}\left(x\right)}{1-x}dx=2\sum_{k\geq0}\int_{0}^{1}\log^{2}\left(x\right)x^{k}dx $$ $$=4\sum_{k\geq1}\frac{1}{k^{3}}=4\zeta\left(3\right) $$ so using the previous result we have $$I=\zeta\left(3\right)\left(2-\frac{2}{5}\right)=\frac{8}{5}\zeta\left(3\right). $$

I'll prove the second integral in the post.

Let $I(a)=\int^{\infty}_{0}\frac{\log y\log(1+ay)}{y(y+1)}dy$, we have $I'(a)=\int^{\infty}_{0}\frac{\log y\,dy}{(1+y)(1+ay)}$, and $I=I(\frac{3+\sqrt{5}}{2})+I(\frac{3-\sqrt{5}}{2})$.

Now $I(0)=0$, and $I'(a)=\frac{(\log a)^2}{2(1-a)}$(Proof: Substitute $y\rightarrow\frac{1}{ay}$), so for any $0<a<1$ we have

$\begin{align*} I(a)&=\int^a_0\frac{(\log b)^2}{2(1-b)}db\\ &=\frac12(\log a)^2\log(1-a)+\log a Li_2(a)-Li_3(a) \end{align*}$

and

$\begin{align*} I(1/a)&=\int^{1/a}_0\frac{(\log b)^2}{2(1-b)}db\\ &=\int^{a}_{+\infty}\frac{(\log b)^2}{2b(1-b)}db\\ &=-\frac12(\log a)^2\log(1-a)+\frac16(\log a)^3-\log a Li_2(a)+Li_3(a) \end{align*}$

therefore $\begin{align*} I(a)+I(1/a)&=-(\log a)^2\log(1-a)+\frac16(\log a)^3-2\log a Li_2(a)+2Li_3(a). \end{align*}$

and we put $a=\frac{3-\sqrt{5}}{2}$ to get $\begin{align*} I&=-(\log \frac{3-\sqrt{5}}{2})^2\log(\frac{1+\sqrt{5}}{2})+\frac16(\log\frac{3-\sqrt{5}}{2})^3-2\log \frac{3-\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2})\\ &=\frac{8}{3}(\log\frac{1+\sqrt{5}}{2})^3+4\log \frac{1+\sqrt{5}}{2} Li_2(\frac{3-\sqrt{5}}{2})+2Li_3(\frac{3-\sqrt{5}}{2}). \end{align*}$

The claim follows from the specific values $Li_2(\frac{3-\sqrt{5}}{2})=\frac{\pi^2}{15}-(\log\frac{1+\sqrt{5}}{2})^2$ and $Li_3(\frac{3-\sqrt{5}}{2})=-\frac{2\pi^2}{15}\log\frac{1+\sqrt{5}}{2}+\frac23(\log\frac{1+\sqrt{5}}{2})^3+\frac{4}{5}\zeta(3)$.