Help?

Hi Max: Here is an online "Integral Calculator" that appears to be able to solve your Integral with many steps as well: https://www.integral-calculator.com/

Good luck.

To compute: $\displaystyle\int^{\infty}_{0}\dfrac{\ln^2\left(z\right)dz}{z^4+z^2+1}$.

I think this does come out by Cauchy, but it's quite heavy.

First you need the poles created by the zeros of $z^4+z^2+1=0$: first put $Z=z^2$ so $Z^2+Z+1=0.$ The roots of this equation are the complex cube roots of unity $\omega_1={1\over2}+i\sqrt{3\over2}$ and $\omega_2=\overline\omega_1=\omega_1^2$.

Therefore the roots of $z^4+z^2+1=0$ are the square roots of $\omega_1,\ \omega_2$. But since $\omega_2^2=\omega_1$and $\omega_1^2=\omega_2$ these four square roots are just $z=\pm\omega_2$ and $z=\pm\omega_1$.

You then have to compute the residues at these 4 poles: it helps that $\omega_1=\exp(2\pi i/3)$ being a cube root of unity has absolute value 1 which simplifies the computation of $\ln(z)$.

Having got the residues you then do the usual contour integration trick round two circles radius R and r joined by a cut up the negative real axis. This very much follows the example given in Wikipedia for the case where the denominator is just $(1+z^2)^2$ given here:

https://en.wikipedia.org/wiki/Contour_integration#Example_5_%E2%80%93_the_square_of_the_logarithm

although I think they make rather a meal of it; in particular I think you can ignore the $i\varepsilon$ in the denominators throughout.

Does this help?