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

Put  $$\frac{x}{1-x}=y$$
 $$I=\int_{0}^{\infty}\ln{y}\ln{(1+3y+y^2)}\frac{dy}{y(y+1)}=\frac{8}{5}\zeta{(3)}$$
 Simple integral at first sight, however I cannot prove that. I would appreciate your help.
 A: 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 $Li_3(\frac{3-\sqrt{5}}{2})=-\frac{2\pi^2}{15}\log\frac{1+\sqrt{5}}{2}+\frac23(\log\frac{1+\sqrt{5}}{2})^3+\frac{4}{5}\zeta(3)$.
A: 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 $$I=\zeta\left(3\right)\left(2-\frac{2}{5}\right)=\frac{8}{5}\zeta\left(3\right).
 $$  
A: I can give two guesses as to why this is the case.
One: The indefinite integral (in both cases) has a large closed form solution in terms of polylogs. 
If you factorise the quadratic $1 + x - x^2$, you can expand the function into four functions of the form 
$$\frac{\ln(x - a) \ln(x - b)}{x},$$
and then use integration by parts, linear substitutions and the definition of the polylogarithm. 
Two: Take the third integral formula for Apery's constant. Let $u = e^x$ and you get 
$$\int_1^{\infty} \frac{(\ln u)^2}{u(u + 1)} du$$
which isn't quite your formula but doesn't look too far off. With a more creative subsitution you might be able to finish the job.
