TITLE: How may we show that $\int_{0}^{\infty}{e^x+e^{-x}-3\over (e^x+e^{-x})^2-1}\cdot \ln(x)\mathrm dx={\pi \ln(2)\over 3\sqrt{3}}?$
QUESTION [7 upvotes]: Consider $(1)$

$$\int_{0}^{\infty}{e^x+e^{-x}-3\over (e^x+e^{-x})^2-1}\cdot \ln(x)\mathrm dx={\pi \ln(2)\over 3\sqrt{3}}\tag1$$

My try:
$x=-\ln(t)$ then $(1)$ becomes
$$\int_{0}^{1}{t^2-3t+1\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag2$$
Split $(2):$
$$\color{blue}{\int_{0}^{1}{\ln(-\ln t)}\mathrm dt}+\int_{0}^{1}{t^2-5t\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag3$$
$$\color{blue}{\gamma}+\int_{0}^{1}{t^2-5t\over 2t+1}\cdot{\ln(-\ln t)}\mathrm dt\tag4$$
Where $\gamma$ is Euler-Mascheroni Constant.
How may we prove $(1)?$

REPLY [8 votes]: The factor $\ln x$ tells me that this is a Frullani integral, hidden with an integration by parts. Indeed (forget the constant)
$$
\int\frac{e^x+e^{-x}-3}{(e^x+e^{-x})^2-1}\,dx
=\frac{2}{\sqrt{3}}\biggl(\arctan\frac{1+2e^x}{\sqrt{3}}-\arctan\frac{1+2e^{2x}}{\sqrt{3}}\biggr).
$$
This primitive will cancel the logaritmic singularities at $0$ and $+\infty$, and you are left with the Frullani integral
$$
\begin{aligned}
\int_0^{+\infty}\frac{1}{x}\frac{2}{\sqrt{3}}&\biggl(\arctan\frac{1+2e^{2x}}{\sqrt{3}}-\arctan\frac{1+2e^{x}}{\sqrt{3}}\biggr)\,dx\\
&=\frac{2}{\sqrt{3}}\biggl(\arctan\frac{1+2e^{+\infty}}{\sqrt{3}}-\arctan\frac{1+2e^0}{\sqrt{3}}\biggr)\ln\frac{2}{1}\\
&=\frac{2}{\sqrt{3}}\Bigl(\frac{\pi}{2}-\frac{\pi}{3}\Bigr)\ln 2
=\frac{\pi\ln 2}{3\sqrt{3}}.
\end{aligned}
$$