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.
This would work:
$$I(a) = \int_0^1 \frac{\ln(ax+1)}{x^2+1} dx \\
I’(a) =\int_0^1 \frac{x}{(x^2+1)(ax+1)} dx \\ \overset{\text{partial fractions}}= \\ \frac{-2\ln |ax+1| +\ln(x^2+1)+2a\tan^{-1} x}{2(a^2+1)} \bigg |_0^1 \\ =-\frac{\ln(a+1)}{a^2+1}+\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}
$$
Integrating from $0$ to $1$, $$I(1)-I(0) = -I(1) +\int_0^1 \left(\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}\right) da$$
Hopefully you can finish.
Best Answer
You can get an exponentially convergent series in the following: \begin{align} \int_0^1 \frac{\log(1+x)}{\log(1-x)} \, {\rm d}x &= \int_0^1 \frac{\log(2-x)}{\log(x)} \, {\rm d}x \\ &=\log(2-x) {\rm li}(x) \Big|_0^1 + \int_0^1 \frac{{\rm li}(x)}{2-x} \, {\rm d}x \\ &=\frac{1}{2} \sum_{n=0}^\infty \frac{1}{2^n}\int_0^1 {\rm li}(x) x^n \, {\rm d}x \\ &=-\sum_{n=0}^\infty \frac{\log(2+n)}{(n+1)\,2^{n+1}} \, . \end{align}