Easier approach to $\int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x$? After playing with this integral for a little bit last night, I eventually resorted to complex analysis to solve it.
Can this be solved without complex analysis?  It feels like there ought to be a way.  If not, is there an easier way with complex analysis?  (I'm still pretty beginner tier at this sort of thing.)
My solution is somewhat involved and it is as follows:
First, get rid of the cosh.
$$
\begin{split}
I &= \int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
&= \frac1{2} \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x} + \mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
 \frac{\mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} &= \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \\
2I &= \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x + \int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x
\end{split}
$$
Next, do some u-subs to make it nicer.
$$\begin{split}
u = \mathrm{e}^{-x} & \qquad \mathrm{d}u = - \mathrm{e}^{-x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x &= \int_{0}^{1} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
u = \mathrm{e}^{x}  & \qquad \mathrm{d}u = \mathrm{e}^{x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x &= \int_{1}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
2I &= \int_{0}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\end{split}
$$
Our contour is the counterclockwise semicircular arc of radius $R > 1$ in the upper half of the complex plane.
$$
\begin{split}
\oint_C \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-R}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{R} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi \\
\lim_{R \rightarrow \infty} \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi &= 0 \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= - \int_0^{\infty} \frac{(-z)^{-2/5}}{1 + (-z)^2} \, \mathrm{d}(-z) \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \left(1 + \mathrm{e}^{-2\pi i/5}\right) \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\end{split}
$$
Finally, take the residue and solve for the original integral.
$$
\begin{split}
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= 2 \pi i \operatorname{Res}_{z = i} \left( \frac{z^{-2/5}}{1+z^2} \right) \\
&= 2 \pi i \left( \frac{i^{-2/5}}{2 i} \right) \\
&= \pi i^{-2/5} \\
\int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \pi \left(\frac{i^{-2/5}}{1 + \mathrm{e}^{-2 \pi i / 5}}\right) \\
&= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
2I &= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
I &= \frac{\pi}{4} \left(\sqrt{5} - 1 \right)
\end{split}
$$

Yet another special-function solution, this time using beta-integrals: $$\mathrm{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\int_0^\infty\frac{y^{a-1}\,dy}{(1+y)^{a+b}}\color{blue}{=\int_0^1\frac{z^{a-1}+z^{b-1}}{(1+z)^{a+b}}\,dz},$$ and your integral reduces to a particular case after $z=\mathrm{e}^{-2x}$: $$I=\frac14\int_0^1\frac{z^{-3/10}+z^{-7/10}}{1+z}\,dz=\frac14\mathrm{B}\left(\frac{7}{10},\frac{3}{10}\right)=\frac{\pi}{4\sin(3\pi/10)}=\frac{\pi}{4\cos(\pi/5)}=\ldots$$
