Integral $\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$

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*}

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, $$\int_0^1\,\frac{x\,\ln\left(\frac{1+x}{1-x}\right)}{\Bigg(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\Bigg)^2}\,\text{d}x=I=-S=\frac{1}{240}\,.$$