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.

This is not an answer.
I do not know if, $\forall a >0$, it could be possible to compute
$$I(a)=\int_0^\infty \frac{\log (x)}{e^x+e^{-x}+a}\,dx$$ But for small values of $a$ we could write
$$I(a)=\sum_{n=0}^\infty (-1)^n a^n \int_0^\infty \left(\frac{e^x}{e^{2 x}+1}\right)^{n+1} \log (x)\,dx$$ and we know the exact values of
$$J_n=\int_0^\infty \left(\frac{e^x}{e^{2 x}+1}\right)^{n+1} \log (x)\,dx$$ The first ones are
$$J_0=\frac{1}{4} \pi  \log \left(\frac{4 \pi ^3}{\Gamma
   \left(\frac{1}{4}\right)^4}\right)\qquad \qquad J_1=\frac{1}{4} \left(\gamma +\log \left(\frac{4}{\pi }\right)\right)$$
$$J_2=\frac{1}{32} \pi  \log \left(\frac{4 \pi ^3}{\Gamma
   \left(\frac{1}{4}\right)^4}\right)-\frac{C}{4 \pi }$$
$$J_3=\frac{1}{48} \left(\frac{7 \zeta (3)}{\pi ^2}+2 \gamma +\log (16)-2 \log (\pi
   )\right)$$
$$J_4=-\frac{320 \pi ^2 C+\psi ^{(3)}\left(\frac{1}{4}\right)-4 \pi ^4 \left(2+9
   \log (4)+27 \log (\pi )-36 \log \left(\Gamma
   \left(\frac{1}{4}\right)\right)\right)}{6144 \pi ^3}$$
$$J_5=\frac{35 \pi ^2 \zeta (3)+93 \zeta (5)+8 \pi ^4 \left(\gamma +\log
   \left(\frac{4}{\pi }\right)\right)}{960 \pi ^4}$$
The next do not fit on a line but they do not make any problem (at least up to $n=7$; for $n>7$ start appearing derivatives of the zeta function).
For $a=1$ and the summation up to $n=7$, the decimal value is $-0.125365$ while the exact value is $-0.126321$.
For $a=0.5$, the decimal value is $-0.179238$ while the exact value is $-0.179243$.
