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. 
 A: Let us define
$$I(a) := \int_0^\infty \frac{\ln(x)}{x^2+a^2}dx = \frac{\pi \ln(a)}{2a}$$
As suggested in the comments by Yuriy S., we can use Leibniz' rule of differentiation under the integral sign here.$^{(1)}$ Take the derivative with respect to $a$ throughout:
$$\begin{align}
I'(a) &= \int_0^\infty \ln(x) \cdot \frac{\partial}{\partial a} \left( \frac{1}{x^2+a^2} \right)dx = -2a \int_0^\infty \frac{\ln(x)}{(x^2+a^2)^2}dx \\
&= \frac \pi 2 \cdot \frac{d}{da} \left( \frac{\ln(a)}{a} \right) = \frac{\pi}{2a^2} \left(1 - \ln(a) \right)
\end{align}$$
The integral you desire is given by $I'(e)$, divided by a factor of $-2e$, but that factor doesn't matter in the end since $I'(e) = 0$ and thus so does your integral.

$^{(1)}$ - Leibniz's rule more generally states the following:
$$ \frac{d}{dx} \int_a^b f(x,t)dt = \int_a^b \frac{\partial}{\partial x} f(x,t)dt$$
when $f$ and $f_x$ are continuous functions (over, if necessary, just some region of the $xt$ plane, e.g. for $t \in (a,b)$ and $x$ in some interval about the point where you intend to evaluate it). I believe it's sufficiently clear that these hold in your case.
A PDF detailing the rule and many examples can be found here.

I don't claim this to be the only way to evaluate the integral. Indeed, that it is equal to zero makes me wonder if there might be some means of transforming it so that you have an odd integrand on a symmetric interval. I'm also not 100% sure if this is the kind of method you wanted to look at given your mentions of complex contour integration.
A: $$\mathcal I:=\int_0^\infty \frac{\ln x}{x^2+e^2}dx=\frac{\pi}{2e}$$
$$\mathcal J:=\int_0^\infty \frac{\ln x}{(x^2+e^2)^2}dx$$
$$\mathcal I-e^2\mathcal J=\int_0^\infty \frac{\ln x}{x^2+e^2}dx-e^2\int_0^\infty \frac{\ln x}{(x^2+e^2)^2}dx=\int_0^\infty \frac{x^2\ln x}{(x^2+e^2)^2}dx$$
$$\overset{IBP}=-\underbrace{\frac12 \frac{x\ln x}{x^2+e^2}\bigg|_0^\infty}_{=0}+\frac12\int_0^\infty \frac{1+\ln x}{x^2+e^2}dx=\frac12\left(\frac{\pi}{2e}+\mathcal I\right)$$
$$\require{cancel}\Rightarrow \cancel{\frac12\mathcal I}-e^2\mathcal J=\cancel{\frac{\pi}{4e}} \Rightarrow \mathcal J=0$$

The result for $\mathcal I$ was used only because your problem asked so, but there is no need for that.
Above we showed that:
$$\int_0^\infty \frac{\ln (x^2)}{x^2+e^2}dx-2e^2\int_0^\infty \frac{\ln x}{(x^2+e^2)^2}dx=\int_0^\infty \frac{\ln(ex)}{x^2+e^2}dx$$
$$\Rightarrow \int_0^\infty \frac{\ln (\frac{x}{e})}{x^2+e^2}dx=2e^2\int_0^\infty \frac{\ln x}{(x^2+e^2)^2}dx$$
$$\int_0^\infty \frac{\ln (\frac{x}{e})}{x^2+e^2}dx\overset{\large x\to\frac{e^2}{x}}=\int_0^\infty \frac{\ln\left(\frac{e}{x}\right)}{x^2+e^2}dx=-\int_0^\infty \frac{\ln (\frac{x}{e})}{x^2+e^2}dx$$
$$\Rightarrow 2\int_0^\infty \frac{\ln (\frac{x}{e})}{x^2+e^2}dx=0\Rightarrow \int_0^\infty \frac{\ln x}{(x^2+e^2)^2}dx=0$$
For those who are not restricted to a specific method, I will also suggest the substitution $x\to\frac{a^2}{x}$ instead of contour integration for the original integral in order to get $\frac{\pi \ln a}{2a}.$
A: Here is a proof without using contour integration ,
$$\int_0^\infty\frac{\ln x}{a^2+x^2}dx\overset{x=at}{=}\frac1a\int_0^\infty\frac{\ln a+\ln t}{1+t^2}dt$$
$$=\frac{\ln a}{a}\underbrace{\int_0^\infty\frac{dt}{1+t^2}}_{\pi/2}+\frac1a\underbrace{\int_0^\infty\frac{\ln t}{1+t^2}dt}_{0}$$
The result of the last integral follows from $t\mapsto 1/t$
