integration - Finding $int_0^infty frac{ln{x}}{(x^2+e^2)^2}dx$

How would one approach this integral?
$$\int_0^{\infty} \frac{\ln{x}}{(x^2+e^2)^2}dx$$

For context, this was a part of an integration exercise, where the previous parts led to showing that $\int_0^{\infty} \frac{\ln{x}}{(x^2+a^2)}dx=\frac{\pi \ln{a}}{2a}$, using a keyhole method involving residues (these parts were fine).

We are required to use this result in solving the integral mentioned at the beginning of this post.

Could someone guide me on how to approach this part? My first thought process was to let $a = e$ in the proven result, then multiplying top and bottom by $(x^2+e^2)$ in order to make it look like the integral we are trying to solve:

$$\int_0^{\infty} \frac{\ln {x}}{(x^2+e^2)}dx = \int_0^{\infty} \frac{x^2 \ln{x}}{(x^2+e^2)^2}dx+\int_0^{\infty}\frac{e^2\ln x}{(x^2+e^2)^2}dx$$

If one uses the $u=e^2/x$ substitution on the first term, then they can introduce the integral given in the question, but this results in having to evaluate the integral of a rational function with a fourth power denominator. I guess one could use partial fractions or contour integrals again to deal with this part, but I have a feeling that this approach probably not the intention of the exercise, although I may be wrong.

Any help, and especially any help with some kind of description of the motivations behind the steps, would be greatly appreciated.