An improper integral with exponentials

Simplification. For every $x\in\mathbb{R}$,
$$\frac{1}{(1+e^x)(1+e^{-x})}=\frac{e^x}{(1+e^x)(e^x+1)}=\frac{e^x}{(1+e^x)^2}.$$
Indeed $1+e^{-x}=\frac{e^x+1}{e^x}$, so $(1+e^x)(1+e^{-x})=\frac{(1+e^x)^2}{e^x}$.

Antiderivative on $[0,a]$. For $a>0$,
$$\int_0^a\frac{e^x}{(1+e^x)^2}\,dx=\left[-\frac{1}{1+e^x}\right]_0^a=-\frac{1}{1+e^a}+\frac{1}{1+e^0}=\frac{1}{2}-\frac{1}{1+e^a}.$$

Improper integral. The integrand is continuous and positive on $[0,+\infty[$, and
$$\int_0^{+\infty}\frac{dx}{(1+e^x)(1+e^{-x})}=\lim_{a\to+\infty}\left(\frac12-\frac{1}{1+e^a}\right) = \frac12.$$