Question
How can we evaluate the following integral?
$$ \mathcal{I} = \int_0^\infty \frac{1}{(1 - x^2 + x^4)} \ln\left({\frac{x^2}{1 - x^2 + x^4}}\right) dx $$
My attempt
Substituting $x^2 = u \implies 2x \cdot dx = du$ gives:
$$ \mathcal{I} = \int_0^\infty \frac{1}{1-u+u^2} \ln\left({\frac{u}{1-u+u^2}}\right) \frac{du}{2\sqrt{u}} $$
Similarly, substituting $\frac{u}{1 - u + u^2} = v$:
$$\frac{1 - u^2}{(1 - u + u^2)^2} \cdot du = dv \implies \frac{du}{1 - u + u^2} = \frac{1 - u + u^2}{1 - u^2} dv$$
So, we have the integral boundaries: from $0 \to \infty$ to $0 \to \lim\limits_{u \to 0} u$. I expressed the new upper limit in limits because the fractional value approaches zero (but not exactly zero) as $v \to 0$ as $u \to \infty$.
This yields:
$$ \mathcal{I} = \int_0^{\lim\limits_{u \to 0} u} \ln v \cdot \frac{1 - u + u^2}{1 - u^2} \cdot \frac{du}{2 \sqrt{u}} $$
From here, I don't know how to proceed.
Notes to the readers
A scientific calculator and integral-calculator.com fails to evaluate this integral.
Wolfram|Alpha yields -2.17759 ($\pi \ln2$).
