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?
 A: Since the thread mentioned in my comment above is closed, I am not sure that you have sufficient reputation to see anything.  I decided to repost my answer, even if this is not the kind of solutions you requested for, and frankly, I do not have any idea how to do the job without contour integration.  The question in that thread is here.

Compute $$S:= \int_{-1}^{+1}\,\frac{x\,\ln(1-x)}{\big(\pi^2+4\,\operatorname{arctanh}^2(x)\big)^2}\,\text{d}x\,.$$

Note that 
$$S=\int_{-1}^{+1}\,\frac{x\,\ln(1-x)}{\Big(\pi^2+4\,\big(\text{arctanh}(x)\big)^2\Big)^2}\,\text{d}x=-\int_{-1}^{+1}\,\frac{x\,\ln(1+x)}{\Big(\pi^2+4\,\big(\text{arctanh}(x)\big)^2\Big)^2}\,\text{d}y$$
That is,
$$S=-\frac12\,\int_{-1}^{+1}\,\frac{x\,\ln\left(\frac{1+x}{1-x}\right)}{\Big(\pi^2+4\big(\text{arctanh}(x)\big)^2\Big)^2}\,\text{d}x=-\,\int_{-\infty}^{+\infty}\,\frac{t\,\tanh(t)\,\big(\text{sech}(t)\big)^2}{\left(\pi^2+4t^2\right)^2}\,\text{d}t$$
where $t:=\text{arctanh}(x)=\frac{1}{2}\,\ln\left(\frac{1+x}{1-x}\right)$.   Note that
$$-S=\int_0^1\,\frac{x\,\ln\left(\frac{1+x}{1-x}\right)}{\Bigg(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\Bigg)^2}\,\text{d}x$$
is the required integral $I$ in Zacky's question.
For each positive integer $N$, consider the contour $C_N$ (oriented in the counterclockwise direction), which is defined to be
$$\small \left[-N^2\pi,+N^2\pi\right]\cup \left[+N^2\pi,+N^2\pi+N\pi\text{i}\right]\cup\left[+N^2\pi+N\pi\text{i},-N^2\pi+N\pi\text{i}\right]\cup\left[-N^2\pi+N\pi,-N^2\pi\right]\,.$$
We note that
$$\lim_{N\to\infty}\,\oint_{C_N}\,f(z)\,\text{d}z=\int_{-\infty}^{+\infty}\,f(t)\,\text{d}t=-S\,,$$
where
$$f(z):=\frac{z\,\tanh(z)\,\big(\text{sech}(z)\big)^2}{\left(\pi^2+4z^2\right)^2}\text{ for all }z\in\mathbb{C}\setminus\Biggl\{\left(n-\frac12\right)\pi\text{i}\,\Big|\,n\in\mathbb{Z}\Biggr\}\,.$$
However,
$$\lim_{N\to\infty}\,\oint_{C_N}\,f(z)\,\text{d}z=2\pi\text{i}\,\sum_{n=1}^\infty\,\text{Res}_{z=\left(n-\frac12\right)\pi\text{i}}\big(f(z)\big)\,.$$
Let $r_n:=\text{Res}_{z=\left(n-\frac12\right)\pi\text{i}}\big(f(z)\big)$ for $n=1,2,3,\ldots$.  Then, with very tedious, but absolutely standard and boring, algebraic manipulations (meaning that I refuse to write my two-page calculations here), we get 
$$r_1=-\frac{3\text{i}}{32\pi^5}-\frac{\text{i}}{480\pi}$$
and, for $n=2,3,4,\ldots$,
$$r_n=\frac{3(2n-1)(2n^2-2n+1)\text{i}}{32\pi^5n^4(n-1)^4}=\frac{3\text{i}}{32\pi^5}\,\left(\frac{1}{(n-1)^4}-\frac{1}{n^4}\right)\,.$$
Ergo,
$$\sum_{n=1}^\infty\,r_n=-\frac{\text{i}}{480\pi}\,.$$
Thus,
$$S=-\int_{-\infty}^{+\infty}\,f(t)\,\text{d}t=-\lim_{N\to\infty}\,\oint_{C_N}\,f(z)\,\text{d}z=-2\pi\text{i}\,\sum_{n=1}^\infty\,r_i=-\frac{1}{240}\,.$$
Therefore,
$$\int_0^1\,\frac{x\,\ln\left(\frac{1+x}{1-x}\right)}{\Bigg(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\Bigg)^2}\,\text{d}x=I=-S=\frac{1}{240}\,.$$
A: 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*}
