How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx$
Using the same method, I couldn't find my asked integral. Are there any ways to connect them? Any help would be appreciated.

The similar problem was posted on AoPS the other day. There are nice answers posted. For the sake of completeness, I would like to add the solution, based on approach developed by Yaroslav Blagouchine; it is convenient for solving the problems with a specific symmetry by means of the integration along a rectangular contour in the complex plane.
Let
$$I=\int_0^\infty \frac{\ln{x}}{e^{x}+e^{-x}+1}dx=\int_0^\infty \frac{\ln{x}}{2\cosh x+1}dx$$
$$=\lim_{a\to0}\frac{1}{2}\int_0^\infty\frac{\ln(x^2+a^2)}{2\cosh x+1}dx=\frac{1}{4}\lim_{a\to0}\Re\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx$$
Let's consider
$$I(a)=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx=\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln(a-2\pi i t)}{\cosh 2\pi t+\frac{1}{2}}dt$$
$$=\frac{\pi}{2}\ln2\pi\int_{-\infty}^\infty\frac{dt}{\cosh 2\pi t+\frac{1}{2}}+\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t+\frac{1}{2}}dt=I_1+I_2(a)$$
Now, we move to the complex plane and consider the closed rectangular contour $-R\,\to R\,\to (R+i)\,\to (-R+i)\,\to -R;\,\,R\to \infty$; counter clockwise.
Let's also consider the following integral along this contour. The integrand has simple poles at the points $z=\frac{i}{3}$ and $\frac{2i}{3}$.
$$\frac{\pi}{2}\ln2\pi\oint\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}dz=I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi \,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}$$
$$I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi\frac{2\pi i}{\sqrt 3\pi i}\Big(e^{-\frac{\beta}{3}}-e^{-\frac{2\beta}{3}}\Big)$$
(We also have to add side integrals - along $R\to R+i$ and $-R+i\to-R$, but these integrals $\to 0$ at $R\to\infty$). Taking the limit $\beta\to 0$, we find
$$\boxed{\,\,I_1(0)=I_1=\frac{\pi}{3\sqrt 3}\ln2\pi\,\,}$$
To evaluate $I_2(a)$, we notice that
$$\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}=\frac{\ln\Gamma\big(\frac{a}{2\pi}-it+1\big)-\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
$$=\frac{\ln\Gamma\big(\frac{a}{2\pi}-i(t+i)\big)}{\cosh 2\pi (t+i)-\frac{1}{2}}\,-\,\frac{\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
(We used the fact that $\cosh (x+2\pi i)=\cosh x$). Adding two integrals (along $R\to R+i$ and $-R+i\to-R$ - these integrals $\to 0$ at $R\to\infty$), we can present $I_2(a)$ in the form of the integral along the same rectangular contour
$$I_2(a)=-\frac{\pi}{2}\oint\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}dz=-\frac{\pi}{2}\,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}=-\frac{\pi}{\sqrt 3}\Big(\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)-\ln\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)\Big)$$
$$\boxed{\,\,I_2(a)=\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)}{\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)}\,\,}$$
Coming back to our initial integral
$$I=I_1+\Re\,I_2(0)=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{2}{3}\big)}{\Gamma\big(\frac{1}{3}\big)}$$
Using the reflection formula for gamma-function $\Gamma\big(\frac{2}{3}\big)\Gamma\big(\frac{1}{3}\big)=\frac{\pi}{\sin\frac{\pi}{3}}=\frac{2\pi}{\sqrt3}$
$$\boxed{\,\,I=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{2\pi}{\sqrt3}-\frac{2\pi}{\sqrt 3}\ln\Gamma\Big(\frac{1}{3}\Big)=-0.126321...\,\,}$$
