TITLE: Evaluate $\int\limits_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx$
QUESTION [6 upvotes]: I was doing experiments with Wolfram Alpha online calculator, about similar integrals (simpler than the below one) and wondered about how get a closed-form for $$\int_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx\tag{1}.$$ 
I've calculated the definite integral using the online calculator, but I believe that the output is an approximation, and since after of this, I've asked to Wolfram Alpha about the indefinite integral, I know that Wolfram Alpha can calculate it, but to me is impossible to evaluate the terms (are about two pages) 
int log(1-x+x^2)log(1+x-x^2)/x dx

Question. Is there some way to evaluate this integral in $(1)$? This was just a curiosity, but I am asking here if you know such integral or do you know how get the evaluation of our integral. Thanks in advance.

Since Wolfram Alpha's answer seems to me difficult, I didn't any attempt (change or variable, integration by parts...).

REPLY [3 votes]: Here's how a physicist is dealing with such a problem.
First of all, note the following relation: 
$$1-x+x^2=\frac{1+x^3}{1+x}$$ 
It allows to split the original integral into two integrals:
$$I_1=\int_0^1\frac{\ln(1+x^3)\ln(1+x-x^2)}{x}dx;(1)$$
$$I_2=-\int_0^1\frac{\ln(1+x)\ln(1+x-x^2)}{x}dx;(2)$$
Even these integrals seem not to have easy manageable closed form solutions.
So we get out of the mainstream.
First step is to note that $y=\ln(1+x)$ satisfies the following differential equation: $$(1+x)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0;(3)$$ 
On the other hand, if we consider the following functional:  
$$J[y(x)]=\int_0^1(1+x)\left ({\frac{dy}{dx}}  \right )^2dx;(4)$$  
then $(3)$ is so called Euler-Lagrange Differential Equation for $(4)$
(See Wikipedia for ex.)
It means that we can use $(4)$ to find an approximate expression for $y=\ln(1+x)$ such that $(1)$ and $(2)$ can be evaluated in closed forms.
As a first approximation we choose a simplest expression: 
$$y_1=x\ln(2)+cx(x-1);(5)$$ 
Next put $(5)$ into $(4)$ to get 
$$J(c)=\frac{c^2}{2}+\frac{c\ln(2)}{3}+\frac{3\ln^2(2)}{2};(6)$$
An extremum point of $(6)$ is $c=-\frac{\ln(2)}{3}$
Put this into $(5)$ to get
$$y_1=\frac{\ln(2)}{3}(4-x)x;(7)$$
That's what we have been looking for, an approximation of $y=\ln(1+x)$ in the range $(0,1).$
For $y=\ln(1+x^3)$ we get from $(7)$:
$$y_1=\frac{\ln(2)}{3}(4-x^3)x^3;(8)$$
Now we are able to evaluate $(1)$ and $(2)$ in closed forms, replacing 
$y=\ln(1+x^3)$ and $y=\ln(1+x)$ in $(1)$ and $(2)$ with $(8)$ and $(7)$ respectively.
I skip routine calculations and write down only the end result:
$$I=I_1+I_2=-\frac{13\ln(3)}{1080}(60\sqrt{5}\ln\frac{3+\sqrt{5}}{2}-119)=-0.0844...$$
Absolute error from exact value is less than $0.00042$. 
Not bad.