Start with the generating functions for the Bernoulli/Euler Polynomials.
$$ \begin{align} &\frac{e^{-ax}}{x^{n}\left(e^{-x}-1\right)}=-\sum_{k=0}^{\infty}\left(-1\right)^{k}B_k\left(a\right)\frac{x^{k-n-1}}{k!},\qquad |x|<2\pi\\ &\frac{2e^{-ax}}{x^{n}\left(e^{-x}+1\right)}=\space\space\space\sum_{k=0}^{\infty}\left(-1\right)^{k}E_k\left(a\right)\frac{x^{k-n}}{k!},\quad\qquad |x|\le\pi \end{align} $$
Integrating from x to $\infty$, we have
$$ \begin{align} &\int_{x}^{\infty}\frac{e^{-at}}{t^{n}\left(e^{-t}-1\right)}dt\\\tag{1} &=C^-_{n}(a)\\ &+\sum_{k=0}^{n-1}\left(-1\right)^{k}B_k\left(a\right)\frac{x^{k-n}}{k!\left(k-n\right)}+\left(-1\right)^{n}B_n\left(a\right)\frac{1}{n!}\ln x+\sum_{k=n+1}^{\infty}\left(-1\right)^{k}B_k\left(a\right)\frac{x^{k-n}}{k!\left(k-n\right)},\qquad |x|<2\pi \\\\\\ &\int_x^\infty\frac{2e^{-at}}{t^n(e^{-t}+1)}dt\\\tag{2} &=C^+_{n}(a)\\ &-\left(\sum_{k=0}^{n-2}\left(-1\right)^{k}E_k\left(a\right)\frac{x^{k-n+1}}{k!\left(k-n+1\right)}+\left(-1\right)^{n-1}E_{n-1}\left(a\right)\frac{1}{\left(n-1\right)!}\ln{x}+\sum_{k=n}^{\infty}\left(-1\right)^{k}E_k\left(a\right)\frac{x^{k-n+1}}{k!\left(k-n+1\right)}\right),\\ &\qquad\qquad |x|\le\pi \end{align} $$
I'm interested in the constants of integration $C^-$ and $C^+$. Some values for $C^+$ are listed below:
$$ \begin{align} C^+_{1}(1)&=\ln\pi-\ln2-\gamma\\ C^+_{1}(2)&=-\ln\pi+\ln2-\gamma\\ C^+_{1}(3)&=\ln\pi-3\ln2-\gamma\\ \\ C^+_{2}(1)&=-6\ln A+\frac{2}{3}\ln2+\frac{1}{2}\gamma\\ C^+_{2}(2)&=6\ln A-\frac{2}{3}\ln2+\frac{3}{2}\gamma-2\\ C^+_{2}(3)&=-6\ln A+\frac{14}{3}\ln2+\frac{5}{2}\gamma-2\\ \\ C^+_{3}(1)&=\frac{7}{4\pi^{2}}\zeta(3)\\ C^+_{3}(2)&=-\frac{7}{4\pi^{2}}\zeta(3)-\gamma+\frac{3}{2}\\ C^+_{3}(3)&=\frac{7}{4\pi^{2}}\zeta(3)-4\ln2-3\gamma+\frac{9}{2}\\ \\ C^+_{4}(1)&=5\zeta'(-3)-\frac{2}{45}\ln2-\frac{1}{24}\gamma+\frac{11}{144}\\ C^+_{4}(2)&=-5\zeta'(-3)+\frac{2}{45}\ln2+\frac{3}{8}\gamma-\frac{11}{16}\\ C^+_{4}(3)&=5\zeta'(-3)+\frac{118}{45}\ln2+\frac{55}{24}\gamma-\frac{605}{144} \end{align} $$
Is there a nice closed form for $C^-$ and $C^+$?
Here's how I calculated the values for $C^+$. It's easy to see that
$$\frac{d}{da}C^+_{n+1}(a)=-C^+_{n}(a)$$
Additionally $E_n(0)=-E_n(1)\implies C^+_{n}(0)=-C^+_{n}(1)$. We can use this to get $$C^+_{n+1}(a)=-\frac12\left(\int_0^aC^+_{n}(v)dv+\int_1^aC^+_{n}(v)dv\right)$$
For the initial value, I had previously found the following for $n=1$.
$$ C^+_{1}(a)=-\ln2-\gamma-2\ln\Gamma\left(\frac{a+1}{2}\right)+2\ln\Gamma\left(\frac{a}{2}\right) $$
For $C^-$ we also have
$$\frac{d}{da}C^-_{n+1}(a)=-C^-_{n}(a)$$ and an equation for $n=1$, $$C^-_{1}(a)=\frac{1}{2}\ln2\pi-\left(a-\frac{1}{2}\right)\gamma-\ln\Gamma(a)$$
However since $B_n(0)=B_n(1)\implies C^-_{n}(0)=C^-_{n}(1)$, we can't use the same trick for $C^-$. So while I have a method for solving $C^+$ I'm not sure how to get anywhere with $C^-$ for $n>1$.
EDIT 5/13/21
I have a working conjecture that
$$C^+_n(a)=2\left(-1\right)^{n+a+1}\frac{2^{n}-1}{\left(n-1\right)!}\zeta'\left(1-n\right)-\left(-1\right)^{a}\frac{2^n}{2^n-1}\frac{E_{n-1}\left(1\right)}{\left(n-1\right)!}\ln2+\left(-1\right)^{n}\frac{\gamma-H_{n-1}}{\left(n-1\right)!}E_{n-1}\left(a\right)+\left(-1\right)^{n+a+1}\frac{2}{\left(n-1\right)!}\sum_{k=1}^{a-1}\left(-1\right)^{k}k^{n-1}\ln k$$
EDIT 9/30/23 I realized I can use $B_n(1/2)=\left(\frac{1}{2^{n-1}}-1\right)B_n(0)$ implying
$$C^-_{n}(1/2)=\left(\frac{1}{2^{n-1}}-1\right) C^-_{n}(0)-\frac{B_n}{2^{n-1}n!}\ln 2$$
to get
$$C^-_{n+1}(a)=\left(\frac{1}{\frac{1}{2^{n-1}}-2}\right)\left(\int_{1/2}^a C^-_{n}(v)dv-\left(\frac{1}{2^{n-1}}-1\right)\int_0^a C^-_{n}(v)dv+\frac{B_n}{2^{n-1} n!}\ln{2}\right)$$
This means we get the next few solutions.
$$ \begin{align} C^-_{2}(a)&=\left(\frac{a^{2}}{2}-\frac{a}{2}+\frac{1}{12}\right)\gamma-\frac{a}{2}\ln\left(2\pi\right)-\ln A+\psi^{(-2)}\left(a\right) \\ C^-_{3}(a)&=-\left(\frac{a^{3}}{6}-\frac{a^{2}}{4}+\frac{a}{12}\right)\gamma+\frac{a^{2}}{4}\ln\left(2\pi\right)+a\ln A+\frac{1}{8\pi^{2}}\zeta(3)-\psi^{(-3)}\left(a\right) \end{align} $$
