<s>[INST] Integral $\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$ The goal is to show (preferably without contour integration, as my knowledge is pretty limited there, but if you can do it that way there is no problem to share it) that: $$I= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}\mathrm dx=\frac{1}{240}$$ 
I am still trying to find more relevant approaches, but so far I thought of: Considering $$J(a)= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{a^2+\ln^2\left(\frac{1+x}{1-x}\right)}dx$$ We have that $J'(\pi)=-2\pi I$, also $$J(a)=\int_0^\infty e^{-at}\int_0^1 x\sin\left(t\ln\left(\frac{1+x}{1-x}\right)\right)dx dt $$ Which is nice, however I couldnt solve the inner integral.
Another approach would be to let $\ln\left(\frac{1+x}{1-x}\right)=t \rightarrow x=\frac{e^t-1}{e^t+1}\rightarrow dx=\frac{2e^t}{(e^t+1)^2}$ 
$$I=2\int_0^\infty \frac{e^t(e^t-1)}{(e^t+1)^3}\frac{t}{(\pi^2+t^2)^2}dt$$ This doesnt look nice, but notice that $$\frac{d^2}{dx^2} \left(\frac{1}{1+e^x}\right)=\frac{e^x(e^x-1)}{(e^x+1)^3}$$ Integrating by parts two times I get $$I=\frac{1}{\pi^2} +24 \int_0^\infty \frac{x^2-\pi^2}{(x^2+\pi^2)^4}\frac{x}{1+e^x}dx$$ Now I am thinking to use series for $\frac{x}{1+e^x}$, but I dont know a series that satisfies it, similary we have for $\frac{x}{e^x-1}$ in terms of Bernoulli numbers. Could you land some help for this integral?
 [/INST] We will use $3$ instances of Schroder's formula which evaluate Gregory coefficients https://en.wikipedia.org/wiki/Gregory_coefficients#Computation_and_representations
\begin{eqnarray*}
-G_2=\int_0^{\infty} \frac{dx}{(1+x)^2(\pi^2+(\ln x)^2)} =\frac{1}{12} \\
G_3=\int_0^{\infty} \frac{dx}{(1+x)^3(\pi^2+(\ln x)^2)} =\frac{1}{24} \\
-G_4=\int_0^{\infty} \frac{dx}{(1+x)^4(\pi^2+(\ln x)^2)} =\frac{19}{720} \\
\end{eqnarray*}
I am not sure how you would show these without using residue calculus ... but try & ignore this $\ddot \smile$
We want to show 
\begin{eqnarray*}
I=\int_0^1 \frac{x \ln (\frac{1+x}{1-x}) } {\left(\pi^2+\left(\ln (\frac{1+x}{1-x}) \right)^2\right)^2 } dx = \frac{1}{240}.
\end{eqnarray*}
Substitute $u=\frac{1-x}{1+x}$ ( so $dx = \frac{-2 du}{(1+u)^2}$) and the intgeral becomes
\begin{eqnarray*}
I=-2\int_0^1 \frac{(1-u) \ln (u) } {(1+u)^3(\pi^2+\left(\ln (u) \right)^2 } du.
\end{eqnarray*}
Now the substitution $v=\frac{1}{u}$ gives exactly the same integrand (but with different limits)
\begin{eqnarray*}
I=-2\int_1^{\infty} \frac{(1-v) \ln (v) } {(1+v)^3(\pi^2+\left(\ln (v) \right)^2 } dv
\end{eqnarray*}
and the integral becomes
\begin{eqnarray*}
I=-\int_0^{\infty} \frac{(1-x) \ln (x) } {(1+x)^3(\pi^2+\left(\ln (x) \right)^2 } dx.
\end{eqnarray*}
Now observe that 
\begin{eqnarray*}
\frac{d}{dx} \frac{1}{\pi^2+(\ln(x))^2} = \frac{-2 \ln(x)}{x(\pi^2+(\ln(x))^2)^2} \\
\end{eqnarray*}
\begin{eqnarray*}
\frac{d}{dx} \frac{x(1-x)} {(1+x)^3} =\frac{x^2-4x+1}{(1+x)^4}
\end{eqnarray*}
so we can integrate by parts to get
\begin{eqnarray*}
I=- \frac{1}{2}\int_0^{\infty} \frac{x^2-4x+1}{(1+x)^4(\pi^2+(\ln(x))^2)}  dx.
\end{eqnarray*}
Rewrite $x^2-4x+1=(x+1)^2-6(x+1)+6$ and use the three results stated at the begining of this answer and we have
\begin{eqnarray*}
I= \frac{1}{2} \left(-6 \frac{19}{720}+6 \frac{1}{24} -\frac{1}{12} \right) = \color{red}{\frac{1}{240}}. 
\end{eqnarray*}
</s>