This integral is a classic example where the order of integration is important.
Your integral can be easily evaluated by rewriting it as
$$\int_0^{\infty}\frac{e^{-y/2}}{y}\Bigg[\int_0^y dx \Bigg]dy$$
Thank you to @Gary for mentioning this term I missed. And here is the full solution.
Use Abel-Plana formula, Eq.(2.10.2)
$$\sum_{n=a}^N f(n)=\int_a^N f(x)dx+\frac{1}{2}f(a)+\frac{1}{2}f(N)+i\int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1}dt+\hat{R}$$
where $\hat{R}$ is the remainder representing the last two terms in the second line of Eq.(2.10.2).
Let $$f(z)=\frac{1}{2}\ln(1+z)$$
then the remainder $\hat{R}$ vanishes when $N\rightarrow\infty$ for our choice of $f(z)$, namely,
$$\lim_{N\to \infty} \hat{R}=0$$
Let
$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt$$
Simplify the following terms
$$f(it)-f(-it)=\frac{\ln(1+it)}{2}-\frac{\ln(1-it)}{2}=\mathrm{artanh}(it)=i\cdot \arctan(t)$$
Let $~a=0$, and we get
$$\sum_{n=0}^N \frac{\ln(1+n)}{2}=\int_0^N \frac{\ln(1+x)}{2}dx+0+\frac{1}{2}\cdot\frac{\ln(1+N)}{2}-\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt+\hat{R}$$
Organize terms and take the limit on both sides,
$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt=\lim_{N\rightarrow \infty} \left( \int_0^N \frac{\ln(1+x)}{2}dx+\frac{\ln(1+N)}{4}-\sum_{n=0}^N \frac{\ln(1+n)}{2} \right) $$
Further, we get
$$I=\frac{1}{2}\lim_{N\rightarrow \infty} \left( ~(N+1)\ln(N+1)-N+\frac{1}{2}\cdot\ln(1+N)- \ln[(N+1)!] ~\right) $$
let $n=N+1$:
$$\begin{align}I&=\frac{1}{2}\lim_{n\rightarrow \infty} \left( ~n\ln(n)-n+1+\frac{1}{2}\cdot\ln(n)- \ln (n!) ~\right)\\
\\
&=\frac{1}{2}\lim_{n\rightarrow \infty} \left(1+\ln\left(\frac{n^{n+\frac{1}2}}{n!e^n}\right)\right) \end{align}$$
Use Stirling's formula, we get the limit:
$$\lim_{n\rightarrow \infty} \left(\frac{n^{n+\frac{1}2}}{n!e^n}\right)=\frac{1}{\sqrt{2\pi}}$$
Finally,
$$\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt =\frac{1}{2}-\frac{1}{4}\cdot\ln(2\pi)$$
Best Answer
Recall Binet's second $\ln \Gamma$ formula:
$$\int_{0}^{\infty} \frac{\arctan (t z)}{e^{2\pi t}-1} \, dt = \frac{1}{2} \ln \Gamma \left(\frac{1}{z}\right) + \frac{1}{2}\left(\frac{1}{z}-\frac{1}{2}\right) \ln (z) + \frac{1}{2z} - \frac{1}{4} \ln (2 \pi)$$
Consider the following integral:
$$\int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz = \frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} + \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)}$$ Since $$\frac{t z}{2 (1+t^2 z^2)(e^{2 \pi t}-1)} = \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right)$$
$$\implies \int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \int_{0}^{\infty} \frac{z}{2} \frac{\partial}{\partial z} \left(\frac{\arctan (t z)}{e^{2 \pi t}-1}\right) \, dt+ \int_{0}^{\infty} \frac{\arctan (t z)}{2 (e^{2 \pi t}-1)} \, dt$$
Using Binet's formula we determine then:
$$\int_{0}^{\infty} \int \frac{t}{(e^{2\pi t}-1)(1+t^2 z^2)^2} \, dz \, dt = \frac{1}{4z}-\frac{1}{8}\ln(2\pi z)-\frac{\psi\left(\frac{1}{z}\right)}{4z}+\frac{1}{4}\ln\left(\Gamma\left(\frac{1}{z}\right)\right)-\frac{1}{8}$$
where $\psi$ is the digamma function.
Taking the derivative with respect to $z$ then the limit as $z \to 1$ we determine:
$$\boxed{\int_{0}^{\infty} \frac{x}{(e^{2 \pi x}-1)(x^2+1)^2} \, dx = \frac{\pi^2}{24} - \frac{3}{8}}$$