<s>[INST] Nice integral $\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$ Last integral of the day :
$$\int_{0}^{\infty}\ln\Big(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\Big)\frac{1}{x}dx=-\frac{3\pi^2}{4}$$
I have tried integration by parts and some obvious substitution but I failed.
I have tried moreover the Feynmann trick without success .WA get an antiderivative  .But I don't see how to get that . I'm not against complex integration if it's necessary but I would like a detailed answer in this case.
So if you have nice ideas...
Thanks a lot for your contributions !
 [/INST] Let $\displaystyle I = \int_0^{\infty} \ln \left( \frac{x^3 - x^2 - x + 1}{x^3 + x^2 + x + 1}\right)\frac{1}{x} dx$. We can write, 
\begin{align}
I & = \int_0^{\infty} \ln \left( \frac{(x-1)^2(x+1)}{(x+1)(x^2 + 1)}\right)\frac{1}{x} dx \\
& = \int_0^{1} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx + \int_1^{\infty} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx \\
& = \int_0^{1} \ln \left( \frac{(x-1)^2}{x^2 + 1}\right)\frac{1}{x} dx + \int_0^{1} \ln \left( \frac{(1-u)^2}{u^2 + 1}\right)\frac{1}{u} du \\
& = 2\int_0^{1} \ln \left( \frac{(1 - x)^2}{x^2 + 1}\right)\frac{1}{x} dx \\
& = 4\int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx - 2\int_0^{1} \frac{\ln \left( 1+ x^2\right)}{x} dx \\
& = 4\int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx - \int_0^{1} \frac{\ln \left( 1+ t\right)}{t} dt
\end{align}
The third step uses the substitution $u = 1/x$ and the last one uses $t = x^2$.
Consider,
\begin{align}
I_1 & = \int_0^{1} \frac{\ln \left( 1 - x\right)}{x} dx \\
& = -\sum_{r = 1}^{\infty} \int_0^1 \frac{x^{r-1}}{r} dx \\
& = -\sum_{r = 1}^{\infty} \frac{1}{r^2} \\
& = -\frac{\pi^2}{6}
\end{align}
Similarly, we can write, $\displaystyle \int_0^{1} \frac{\ln \left( 1+ x\right)}{x} dx = \frac{\pi^2}{12}$.
Plugging the values, we get that $\displaystyle I = -\frac{3 \pi^2}{4}$.
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