A couple of definite integrals related to Stieltjes constants In a (great) paper "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" by Iaroslav V. Blagouchine, the following integral representation of the first Stieltjes constant $\gamma_1$ is given (on page 539):
$$\gamma_1=-\left[\gamma-\frac{\ln2}2\right]\ln2+i\int_0^\infty\frac{dx}{e^{\pi x}+1}\left\{\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\right\}.\tag1$$
It's possible to get rid of imaginary numbers in this formula, and rewrite it in terms of only real-valued functions:
$$2\int_0^\infty\frac{\arctan x}{1+x^2}\frac{dx}{e^{\pi x}+1}-\int_0^\infty\frac{x\ln(1+x^2)}{1+x^2}\frac{dx}{e^{\pi x}+1}=\gamma_1+\left[\gamma-\frac{\ln2}2\right]\ln2.\tag2$$

Question:
  Is it possible to find closed forms separately for each integral on the left-hand side of $(2)$?


Using contour integration, I am able to evaluate the first integral in terms of a slightly different infinite series which may or may not be able to be evaluated in closed form.
First notice that 
$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}  &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \left(1- \tanh \left(\frac{\pi x}{2} \right) \right) \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \, dx- \frac{1}{4} \int_{-\infty}^{\infty} \frac{\arctan x}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \int_{-\infty}^{\infty} \frac{\log(1-ix)}{1+x^{2}} \tanh \left(\frac{\pi x}{2} \right) \, dx  . \end{align} $$
Now let $ \displaystyle f(z) = \frac{\log(1-iz)}{1+z^{2}} \tanh \left(\frac{\pi z}{2} \right).$
Then since $ \displaystyle \int f(z) \ dz$ vanishes along the upper half of the circle $|z|=2N$ as $N \to \infty$ through the positive integers, we get
$$ \begin{align} &\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}  \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left(\text{Res}[f(z),i] + \sum_{n=1}^{\infty} \text{Res} [f(z), i(2n+1)] \right) \\ &= \frac{\pi^{2}}{16} + \frac{1}{4} \, \text{Im} \ 2 \pi i \left( \frac{\log(2)-1}{2 \pi} - \sum_{n=1}^{\infty}\frac{\log(2n+2)}{2 \pi(n^{2}+n)}\right) \\ &= \frac{\pi^{2}}{16} + \frac{\log (2)}{4} - \frac{1}{4} - \frac{\log (2)}{4} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} \\ &= \frac{\pi^{2}}{16} - \frac{1}{4} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} . \end{align}$$
EDIT:
In this answer, Olivier Oloa talks about the coefficients of the regular part of the Laurent series expansion at the origin of the poly-Hurwitz zeta function (which he refers to as the poly-Stieltjes constants).
It turns out that we can evaluate the above infinite series in terms of the poly-Stieltjes constants.
According to Theorem 2, 
$$ \sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)} = \gamma_{1}(1,0) - \gamma_{1}(1,1)$$  which, if I understand correctly, reduces to $$\gamma_{1}(1,0) - \gamma_{1}.$$
As a side note, Wolfram Alpha does not return a very good approximation of the value of that infinite series. And if you ask for more digits, it will return a different result.
But it appears that the value of the infinite series is approximately $ 1.2577468869$.
