Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $? After reading the fantastic solutions to the integral in the post,
$$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x=-\frac{\pi^2}8,$$
and found that $$
\int_0^1 \frac{\ln ^2 x}{1-x^2} d x =\frac{7\zeta(3)}{4}.  $$
Then I keep on exploring the integral in general and obtain a formula,

$$
\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]
$$
where $n\ge 2.$

First of all, we use its partner integral
$$
I(a)=\int_0^1 \frac{x^a}{\left(1-x^2\right)^2} d x
$$
Using the series for $|x|<1,$
$$
\frac{1}{\left(1-x^2\right)^2}=\sum_{k=1}^{\infty} k x^{2 k-2},
$$
we have $$
I(a) =\sum_{k=1}^{\infty} k \int_0^1 x^{2 k-2+a} d x =\sum_{k=1}^{\infty} \frac{k}{2 k-1+a}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln ^n x}{\left(1-x^2\right)^2} d x & =\left.\sum_{k=1}^{\infty} \frac{\partial^n}{d a^n}\left(\frac{k}{2 k-1+a}\right)\right|_{a=0} \\
& =(-1)^n n ! \sum_{k=1}^{\infty} \frac{k}{(2 k-1)^{n+1}} \\
& =\frac{(-1)^n n !}{2} \sum_{k=1}^{\infty}\left[\frac{1}{(2 k-1)^n}+\frac{1}{(2 k-1)^{n+1}}\right] \\
& =\boxed{\frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]}
\end{aligned}
$$
For examples,
$$
I_2=\frac{3}{4} \zeta(2)+\frac{7}{8} \zeta(3)= \frac{\pi^2}{8} +\frac{7}{8} \zeta(3)
$$
$$
I_3=-3\left[\frac{7}{8} \zeta(3)+\frac{15}{16} \zeta(4)\right]=-\frac{21}{8} \zeta(3)-\frac{\pi^2}{32}
$$
My question: Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $?

Utilize
\begin{align}
J(m)=\int_0^1 \frac{\ln^m x}{1-x^2}dx 
=&\int_0^1  \frac{\ln^m x }{1-x}dx -\int_0^1 {\frac{x\ln^m x }{1-x^2} } \overset{x^2\to x}{dx} \\=& \ \left(1-\frac1{2^{m+1}}\right)\int_0^1\frac{\ln^mx}{1-x}dx\\
=&\  \left(1-\frac1{2^{m+1}}\right)(-1)^m m!\ \zeta(m+1)
\end{align}
to evaluate
\begin{align}
\int_0^{1} \frac{\ln^n x}{\left(1-x^2\right)^2}dx
&=\int_0^{1} \frac{\ln ^n x}{2x}\ d\left(\frac{x^2}{1-x^2}\right)
\overset{ibp}=\frac12J(n)
-\frac n2 J(n-1)\\
&= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]
\end{align}
