How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x}  \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't have an explicit closed-form.


An integral relation:
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=-\frac{3}{2}J(k)+\frac{\pi^2}{8k} 
\left ( \operatorname{Li}_2\left ( \frac{1}{k}  \right ) 
-\operatorname{Li}_2\left ( -\frac{1}{k}  \right )  \right )
+\frac{\ln k}{8k}  \ln^3\left ( \frac{k-1}{k+1}  \right ),\qquad (k>1).$$



With some calculations, we followed that
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=\frac{\pi^4}{64k} 
+\frac{3}{4k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1}  \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1}  \right )
  \right ) +\frac{3\pi^2}{8k} \operatorname{Li}_2\left (- \frac{k-1}{k+1}  \right ),\qquad(k>1).$$
Then the final result of $J(k)$ is
$${\color{Green}{J(k)
=\frac{\pi^2}{12k} 
\left ( \operatorname{Li}_2\left ( \frac{1}{k}  \right ) 
-\operatorname{Li}_2\left ( -\frac{1}{k}  \right )  \right )
+\frac{\ln k}{12k}  \ln^3\left ( \frac{k-1}{k+1}  \right )
-\frac{\pi^4}{96k} -\frac{1}{2k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1}  \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1}  \right )
  \right ) -\frac{\pi^2}{4k} \operatorname{Li}_2\left (- \frac{k-1}{k+1}  \right ),\qquad (k>1).}}$$


A solution idea of calculating the integral by Cornel Ioan Valean
The post is extremely short since I have no time, but once you know what to do, all is trivial. So, what do to? Observe that any $J(2n,k)$ is half the real part of the integral over the positive real line.
$$\int_{0}^{1} \frac{\log(x)^{2n}\log\left(\frac{1-x}{1+x}\right)}{(x-1)^2-k^2(x+1)^2}\text{d}x=\frac{1}{2}\Re\biggr\{\int_{0}^{\infty} \frac{\log(x)^{2n}\log\left ( \frac{1-x}{1+x}  \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x\biggr\},$$
which is easy to reach by also using a simple substiution.
Once you have reached this point, exploit multiple integrals involving PV integrals as in the calculations of $J(s)$ here https://math.stackexchange.com/q/3488566.
The rest is easy, boring work to do.
End of story
