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.

$$I=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{x\to 1-x}=\int_0^1 \frac{\ln(1+x-x^2)}{1-x}dx$$
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:
$$I(a)=\int_0^1 \frac{\ln(1+a(x-x^2))}{x-x^2}dx\Rightarrow I'(a)=\int_0^1 \frac{1}{1+a(x-x^2)}dx$$
$$=\frac1a\int_0^1 \frac{1}{\left(\frac1a+\frac14\right)-\left(x-\frac12\right)^2}dx=\frac{2}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)$$
$$I(0)=0\Rightarrow I=\frac12  (I(1)-I(0))=\int_0^1 \frac{1}{\sqrt{a(4+a)}}\ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)da$$
$$\text{let  } \ln\left(\frac{\sqrt{\frac{4+a}{a}}+1}{\sqrt{\frac{4+a}{a}}-1}\right)=x\Rightarrow \frac{1}{\sqrt{a(4+a)}}da=dx$$
$$\Rightarrow I=\int_0^{\ln(\varphi^2)} x dx=\frac12 \ln^2( \varphi^2)=2\ln^2 \varphi$$
