Here is another method which is a mixture of both real-analysis technique and complex-analysis technique (without contour integration). The main idea is similar to my previous answer.
Let $I$ denote the integral and we write
$$ I = \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} x \log(1+x^2) e^{-2\pi n x} \, dx. $$
In order to proceed, we claim the following:
$$ \int_{0}^{\infty} x \log(1+x^2)e^{-sx} \, dx
= \frac{2}{s^2} - \frac{2}{s} \int_{0}^{\infty} \frac{\sin(st)}{t+1} \, dt + \frac{2}{s^2} \int_{0}^{\infty} \frac{\cos(st)}{t+1} \, dt. \tag{1} $$
We postpone the proof to the end and discuss the consequence of $\text{(1)}$. Substituting $s = 2\pi n$ and plugging back, $I$ can be simplified as
\begin{align*}
I
&= \sum_{n=1}^{\infty} (-1)^{n-1} \left( \frac{1}{2\pi^2 n^2} - \frac{1}{\pi n} \int_{0}^{\infty} \frac{\sin(2\pi n t)}{t+1} \, dt + \frac{1}{2\pi^2 n^2} \int_{0}^{\infty} \frac{\cos(2\pi n t)}{t+1} \, dt \right) \\
&= \frac{1}{24} - \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(2\pi n t)}{\pi n} \right) \, \frac{dt}{t+1} + \int_{0}^{\infty} \left( \sum_{n=1}^{\infty}(-1)^{n-1} \frac{\cos(2\pi n t)}{2\pi^2 n^2} \right) \, \frac{dt}{t+1}
\end{align*}
Then utilizing the Fourier series of the periodic Bernoulli polynomials $\tilde{B}_n$
$$ \sum_{n=1}^{\infty} \frac{\sin(2\pi n x)}{\pi n} = -\tilde{B}_1(x), \qquad
\sum_{n=1}^{\infty} \frac{\cos(2\pi n x)}{\pi^2 n^2} = \tilde{B}_2(x), $$
it follows that
$$ I = \frac{1}{24} - \int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{t+1} \, dt - \frac{1}{2}\int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{t+1} \, dt. $$
Next, we introduce the following function
$$f(s) = 2^s - (2^s - 1)\zeta(s) + \frac{1}{s-1}.$$
This is an entire function on $\Bbb{C}$ since all the poles are cancelled out. Then we claim that for $\Re(s) > 0$,
\begin{align*}
\int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{(t+1)^s} \, dt
&= \frac{f(s-1)}{s-1}, \tag{2} \\
\int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{(t+1)^s} \, dt
&= \frac{2f(s-2)}{(s-1)(s-2)} - \frac{1}{12}\frac{1}{s-1}. \tag{3}
\end{align*}
We also postpone the proof to the end. Assuming this, we have
\begin{align*}
\int_{0}^{\infty} \frac{\tilde{B}_1(t + \frac{1}{2})}{t+1} \, dt
&= \lim_{s\to 1} \frac{f(s-1)}{s-1} = \frac{3}{2}\log 2 - 1, \\
\int_{0}^{\infty} \frac{\tilde{B}_2(t + \frac{1}{2})}{t+1} \, dt
&= \lim_{s\to 1} \left(\frac{2f(s-2)}{(s-1)(s-2)} - \frac{1}{12}\frac{1}{s-1}\right) = \frac{7}{12} - \frac{13}{12}\log 2 - \zeta'(-1).
\end{align*}
Plugging this back yields the same answer as M.N.C.E.'s:
$$ I = \frac{3}{4} - \frac{23}{24}\log 2 + \frac{1}{2}\zeta'(-1). $$
Proof of $\text{(1)}$. By the integration by parts, for $s > 0$ we have
\begin{align*}
\int_{0}^{\infty} x \log(1+x^2)e^{-sx} \, dx
&= \int_{0}^{\infty} \frac{2x(sx+1)}{s^2(x^2+1)} e^{-sx} \, dx \\
&= \int_{0}^{\infty} \frac{2u(u+1)}{s^2(u^2 + s^2)}e^{-u} \, du \qquad (u = sx) \\
&= \int_{0}^{\infty} \left( \frac{2}{s^2} - \frac{2}{u^2+s^2} + \frac{2u}{s^2(u^2 + s^2)} \right) e^{-u} \, du.
\end{align*}
In order to compute the last integral, we notice that
$$ \int_{0}^{\infty} \frac{e^{-u}}{u - is} \, du
= \int_{0}^{\infty}\int_{0}^{\infty} e^{ist} e^{-ut} e^{-u} \, dt du
= \int_{0}^{\infty} \frac{e^{ist}}{t+1} \, dt. $$
In fact, this is more of a heuristics than a rigorous computation, since the Fubini's theorem does not apply directly. One can regularize both sides by replacing $s$ by $s - i\epsilon$ for $\epsilon > 0$, applying Fubini's theorem to prove the corresponding equailty, and then letting $\epsilon \to 0^+$ to establish this. (Alternatively, this can be also thought as a $\frac{\pi}{2}$-rotation of the contour.) Then the claim follows by utilizing this equality. ////
Proof of $\text{(2)}$ and $\text{(3)}$. We first prove the following identity
$$ \int_{0}^{\infty} \frac{1}{(t+1)^s} \, d\tilde{B}_1(t+\tfrac{1}{2}) = f(s), \qquad \Re(s) > 0. $$
By integration by parts, it is not hard to check that the left-hand side defines a holomorphic function for $\Re(s) > 0$. In view of the principle of analytic continuation, it suffices to prove the identity for $\Re(s) > 1$. Then
\begin{align*}
\int_{0}^{\infty} \frac{1}{(t+1)^s} \, d\tilde{B}_1(t+\tfrac{1}{2})
&= \int_{0}^{\infty} \frac{1}{(t+1)^s} \, d(t - \lfloor t+\tfrac{1}{2}\rfloor) \\
&= \frac{1}{s-1} - \sum_{k=0}^{\infty} \frac{1}{(k+\frac{3}{2})^s}.
\end{align*}
It is easy to check that this coincides with $f(s)$. Now both $\text{(2)}$ and $\text{(3)}$ follows easily. Again, both define holomorphic functions on $\Re(s) > 0$ and the principle of analytic continuation allows us to prove both identities only when $\Re(s)$ is large. Then the claim follows from the following identity:
\begin{align*}
\int_{0}^{\infty} \frac{\tilde{B}_n(t + \frac{1}{2})}{(t+1)^s} \, dt
&= \frac{B_n(\frac{1}{2})}{s-1} + \frac{1}{s-1} \int_{0}^{\infty} \frac{1}{(t+1)^{s-1}} \, d\tilde{B}_n(t + \tfrac{1}{2}) \tag{$n \geq 1$} \\
&= \frac{B_n(\frac{1}{2})}{s-1} + \frac{n}{s-1} \int_{0}^{\infty} \frac{\tilde{B}_{n-1}(t + \frac{1}{2})}{(t+1)^{s-1}} \, dt \tag{$n \geq 2$}
\end{align*}
