Note that $a>0$, thus I'm not sure if we can apply residues here. (For $a=0$ the integral doesn't converge).
$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}$$
Despite the simple expression under the integral, I didn't find it in G-R book.
I attempted the most straightforward way - geometric series - since $e^x \geq 1$.
$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \int_0^\infty \frac{e^{-kx}dx}{a+x}$$
The integrals on the right can be expressed using incomplete gamma function:
$$\int_0^\infty \frac{e^{-kx}dx}{a+x}=\frac{e^{ka}}{k} \Gamma(0,ka)$$
We obtain:
$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \frac{e^{ka}}{k} \Gamma(0,ka)$$
This is correct, but not very useful, since each incomplete gamma just disguises an integral. It can be computed independently by:
$$\Gamma (0,t)=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$
Thus I suppose we can express the integral as:
$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \cfrac{1}{ka+1-\cfrac{1}{ka+3-\cfrac{4}{ka+5-\cfrac{9}{ka+7-\cdots}}}}$$
Amusing, but again, not very useful.
Does this integral have a closed form? If so, what is it and how do we find it?
Edit
Another interesting expression for incomplete gamma from DLMF:
$$\Gamma (0,t)=e^{-t} \sum_{n=0}^\infty \frac{L_n (t)}{n+1}$$
Here $L_n (t)$ are Laguerre polynomials. This series converges extremely slowly (and oscillates), so it seem quite useless for computation, however gives a pretty expression for the integral:
$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \sum_{n=0}^\infty \frac{L_n (ka)}{n+1} $$
Note, that the continued fraction above converges much faster than the series.
Since: $$\mathcal{L}\left(\frac{1}{1+e^x}\right) = \frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\qquad \mathcal{L}^{-1}\left(\frac{1}{a+x}\right)=e^{-as}$$ the original integral equals: $$ \int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds=\frac{1}{2}\int_{0}^{+\infty}\int_{\frac{s+1}{2}}^{\frac{s+2}{2}}\sum_{n\geq 0}\frac{e^{-as}}{(n+u)^2}\,du\,ds $$ and by integration by parts, the LHS just depends on: $$ \int_{0}^{+\infty}\left[\log\Gamma\left(\frac{s+2}{2}\right)-\log\Gamma\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds$$ that can be further expanded by exploiting the Weierstrass product for the $\Gamma$ function, leading to a fast-convergent series, whose first term is $-\frac{\gamma}{2a}$.