Show that the following integral is convergent
$ \int_{0}^{∞} \log (1+2\operatorname{sech} x) dx $
I tried by replacing $\operatorname{sech}x$ with $ 2/ e^x + e^{-x} $
and then by using limit comparison test with $ g(x) = 1/x^2 $ But couldn't solve further.
$$ 0\leq\int_{0}^{+\infty}\log\left(1+2\operatorname{sech}x\right)\,dx \leq 2\int_{0}^{+\infty}\operatorname{sech}(x)\,dx =\pi $$ since $0\leq \log(1+z)\leq z$ for any $z\geq 0$. Actually
$$ \int_{0}^{+\infty}\log\left(1+2\operatorname{sech}x\right)\,dx =\frac{\pi^2+4\operatorname{arccosh}(2)^2}{8}\approx 2.1008896. $$
Quite simple with equivalents: the positive function $$\operatorname{sech}x\sim_{+\infty}\frac 2{\mathrm{e}^x}$$ and the integral $\;\displaystyle\int_0^{+\infty}\frac 2{\mathrm{e}^x}\,\mathrm d x$ is convergent.
