<s>[INST] Show $\int_{0}^{1}\frac {x^2\ln x }{{(1-x^2)}{(1+x^4)}}dx=\frac{-π^2}{16(2+\sqrt{2})}$ Question:

Prove that $$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-π^2}{16(2+\sqrt{2})}$$

Using partial fraction,we get
\begin{align}
&\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx\\
= &\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1-x}dx+\frac{1}{4}\int_{0}^{1}\frac{\ln x}{1+x}dx+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx
\end{align}
I got first integral as 
$\frac{-\pi^2}{6}$ and second integral as $\frac{-\pi^2}{12}$
$$\int_{0}^{1}\frac {x^2\ln x}{{(1-x^2)}{(1+x^4)}}dx=\frac{-\pi^2}{16}+\frac{1}{2}\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx$$
I got stuck with third integral, which seems difficult to evaluate.
${}{}{}$
A note is also written saying that:-
The reader should evaluate the family of integrals  ${I_{n}=\int_{0}^{1}\frac {x^{2n}\ln x}{{(1-x^2)}{(1+x^4)^n}}dx{,n} \in N}$
The computation of the first few special values indicates an interesting arithmetic structure of the answer.
How to evaluate integral for $n$?
 [/INST] Rewrite the third integral as
$$I=\int_{0}^{1}\frac{(x^2-1)\ln x}{1+x^4}dx 
= -\int_{0}^{\infty}\frac{\ln x}{1+x^4}dx
$$
Use the general result 
$$J(a)= \int_{0}^{\infty}\frac{x^{a-1}}{1+x^m}dx=\frac{\pi}m\csc\frac{\pi a}m
$$
and let $m=4$ to obtain
$$I= - J’(1)= \frac{\pi^2}{8\sqrt2} $$
</s>