Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms. 
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.  

This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post is someone said that this integral cannot be solved without using special functions.
Another alternative solutions will be appreciated.
 A: $$\begin{aligned}
I&=\int_0^1\frac{\ln(1+x-x^2)}x\mathrm{d}x\\
&\overset{(1)}{=}\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n-1}(x-x^2)^n}{nx}\mathrm{d}x\\
&\overset{(2)}{=}\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\int_0^1x^{n-1}(1-x)^n\mathrm{d}x\\
&\overset{(3)}{=}\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\frac{(n-1)!n!}{(2n)!}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(n!)^2}{(2n+2)!}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(1\times2\times\cdots\times n)(1\times2\times\cdots\times n)}{1\times2\times\cdots\times (2n+2)}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(1\times2\times\cdots\times n)}{1\times3\times5\times\cdots\times(2n+1)\times (2n+2)2^n}\\
&=\sum_{n=0}^\infty\frac{(-1)^nn!}{(2n+1)!!(2n+2)2^n}
\end{aligned}$$
Explanation
(1) Using the Maclaurin series of $\ln(1+w)$, where $w=x-x^2$.
(2) It is legal to change the position of $\sum$ and $\int$.
(3) Integrate by parts $n-1$ times.
Notice that $$\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}=\frac{\arcsin x}{\sqrt{1-x^2}},$$
integrate both sides from $0$, we have $$\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!(2n+2)}x^{2n+2}=\frac12\arcsin^2x.$$
Letting $x=\frac i2$ leads to
$$\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n)!!}{(2n+1)!!(2n+2)}2^{-2n-2}=\frac12\arcsin^2\frac i2.$$
Combining with $(2n)!!=2^{n}n!$, we have $$-\frac14I=-\frac12\operatorname{arccsch}^22,$$
or $I=2\ln^2\varphi,$ where $\varphi$ denotes the golden ratio.
A: Put
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{x}\,\mathrm{d}x = \int_{0}^1\dfrac{\ln(1+x(1-x))}{x}\mathrm{d}x.
\end{equation*}
If we change $x$ to $ 1-x $ we get
\begin{equation*}
I=\int_{0}^1\dfrac{\ln(1+x-x^2)}{1-x}\,\mathrm{d}x.
\end{equation*}
Consequently
\begin{equation*}
2I = \int_{0}^1\ln(1+x-x^2)\left(\dfrac{1}{x}+\dfrac{1}{1-x}\right)\,\mathrm{d}x.
\end{equation*}
The next step will be integration by parts.
\begin{equation*}
2I= \underbrace{\left[\ln(1+x-x^2)\ln\dfrac{x}{1-x}\right]_{0}^{1}}_{=0} -\int_{0}^1\dfrac{1-2x}{1+x-x^2}\ln\dfrac{x}{1-x}\, \mathrm{d}x
\end{equation*}
Then\begin{equation*}
I=\dfrac{1}{2}\int_{0}^1\dfrac{2x-1}{1+x-x^2}\ln\dfrac{x}{1-x}\, \mathrm{d}x.
\end{equation*}
If we substitute $ z=\dfrac{x}{1-x} $ we get
\begin{equation*}
I = \int_{0}^{\infty}\dfrac{(z-1)\ln z}{2(z+1)(z^2+3z+1)}\,\mathrm{d}z.
\end{equation*}
In order to evaluate this integral we integrate $\displaystyle \dfrac{(z-1)\log^2 z}{2(z+1)(z^2+3z+1)}$ along a keyhole contour and use residue calculus. We get that
\begin{equation*}
I = 2\ln^2\varphi
\end{equation*}
where $ \varphi = \dfrac{1+\sqrt{5}}{2}. $
A: $$I=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{x\to 1-x}=\int_0^1 \frac{\ln(1+x-x^2)}{1-x}dx$$
Averaging the two integrals from above  gives us:
$$I=\frac12 \int_0^1 \frac{\ln(1+x-x^2)}{x-x^2}dx=\frac12I(1)$$
Where we considered, in order to apply Feynman's trick, the following integral:
$$I(a)=\int_0^1 \frac{\ln(1+a(x-x^2))}{x-x^2}dx\Rightarrow I'(a)=\int_0^1 \frac{1}{1+a(x-x^2)}dx$$
$$=\frac1a\int_0^1 \frac{1}{\left(\frac1a+\frac14\right)-\left(x-\frac12\right)^2}dx=\frac{2}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)$$
$$I(0)=0\Rightarrow I=\frac12  (I(1)-I(0))=\int_0^1 \frac{1}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)da$$
$$\text{let  } \ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)=x\Rightarrow \frac{1}{\sqrt{a(4+a)}}da=dx$$
$$\Rightarrow I=\int_0^{\ln(\varphi^2)} x dx=\frac12 \ln^2( \varphi^2)=2\ln^2 \varphi$$
A: Let $I(a)=\int_0^1\frac{\ln[1+4 x(1-x) \sinh^2a]}xdx$
\begin{align}
I’(a)&= \int_0^1\frac{4(1-x)\sinh2a }{1+4 x(1-x) \sinh^2a}dx\\
&= \left[ \frac{\ln[1+4 x(1-x) \sinh^2a]}{\tanh a}
+2\tanh^{-1}\frac{2x-1}{\coth a}\right]_0^1=4a
\end{align}
Then
$$\int_0^1\frac{\ln(1+x-x^2)}xdx
=\int_0^{\sinh^{-1}\frac12}I’(a)da
= \int_0^{\ln\phi}4a\ da=2\ln^2\phi
$$
A: No need multiple-integrals, no need Feynman's trick
\begin{align}J&=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{\text{IBP}}=-\int_0^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&=-\int_0^{\frac{1}{2}} \frac{(1-2x)\ln x}{1+x(1-x)}dx-\int_{\frac{1}{2}}^1 \frac{(1-2x)\ln x}{1+x(1-x)}dx\\
&\overset{u=x(1-x)}=-\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1-\sqrt{1-4u}}{2}\right)}{1+u}du+\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1+\sqrt{1-4u}}{2}\right)}{1+u}du\\
&=-\int_0^{\frac{1}{4}}\frac{\ln\left(\frac{1-\sqrt{1-4u}}{1+\sqrt{1-4u}}\right)}{1+u}du\overset{z=\sqrt{1-4u}}=2\int_0^1 \frac{z\ln\left(\frac{1-z}{1+z}\right)}{z^2-5}dz\\
&=\int_0^1 \frac{\ln\left(\frac{1-z}{1+z}\right)}{z-\sqrt{5}}dz+\int_0^1 \frac{\ln\left(\frac{1-z}{1+z}\right)}{z+\sqrt{5}}dz\\
&\overset{x=\frac{1-z}{1+z}}=-2\int_0^1 \frac{\ln x}{(1+x)\left((\sqrt{5}+1)x+\sqrt{5}-1\right)}+2\int_0^1 \frac{\ln x}{(1+x)\left((\sqrt{5}-1)x+\sqrt{5}+1\right)}\\
&=-\frac{\sqrt{5}+1}{\sqrt{5}-1}\int_0^1 \frac{\ln x}{\frac{\sqrt{5}+1}{\sqrt{5}-1}x+1}dx-\frac{\sqrt{5}-1}{\sqrt{5}+1}\int_0^1 \frac{\ln x}{\frac{\sqrt{5}-1}{\sqrt{5}+1}x+1}dx+2\int_0^1 \frac{\ln x}{1+x}dx
\end{align}
Let $\rho=\frac{\sqrt{5}+1}{\sqrt{5}-1}$,
\begin{align}J&=\underbrace{-\rho\int_0^1 \frac{\ln x}{\rho x+1}dx}_{u=\rho x}-\underbrace{\frac{1}{\rho}\int_0^1 \frac{\ln x}{\frac{x}{\rho}+1}dx}_{u=\frac{1}{\rho}}+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-\int_0^\rho \frac{\ln\left(\frac{u}{\rho}\right)}{1+u}du-\int_0^{\frac{1}{\rho}} \frac{\ln\left(\rho u\right)}{1+u}du+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-\int_0^\rho \frac{\ln u}{1+u}du-\int_0^{\frac{1}{\rho}} \frac{\ln u}{1+u}du+\ln^2\rho+2\int_0^1 \frac{\ln x}{1+x}dx\\
&=\int_\rho^1 \frac{\ln u}{1+u}du-\underbrace{\int_1^{\frac{1}{\rho}} \frac{\ln u}{1+u}du}_{z=\frac{1}{u}}+\ln^2\rho\\
&=\int_\rho^1 \frac{\ln u}{1+u}du+\int_\rho^1\frac{\ln z}{z(1+z)}+\ln^2\rho\\
&=\int_\rho^1\frac{\ln u}{u}du+\ln^2\rho=\left[\frac{\ln^2 u}{2}\right]_\rho^1+\ln^2\rho=\frac{1}{2}\ln^2\rho\\
\end{align}
Moreover,
Let $\varphi=\dfrac{\sqrt{5}+1}{2}$, then,
\begin{align}\rho=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}+1\right)^2}{4}=\varphi^2\end{align}
Therefore,
\begin{align}J=\frac{1}{2}\ln^2\rho=\boxed{2\ln^2\varphi}\end{align}
