How do I solve this ominous integral? Let $ n\ge 1 $ be a positive integer. How do I prove the generalization: $$ \int_0^1\frac{\arctan(x)\log^{2n}(x)}{1+x} \, dx =\frac{\pi}{4}\left(1-2^{-2 n}\right) \zeta(2 n+1)(2 n)!+\frac{1}{2} \beta(2 n+2)(2 n) !-\frac{\pi}{16} \lim _{s \rightarrow 0}\left(\frac{d^{2 n}}{d s^{2 n}}\left(\csc \left(\frac{\pi s}{2}\right)\left(\psi\left(\frac{3}{4}-\frac{s}{4}\right)-\psi\left(\frac{1}{4}-\frac{s}{4}\right)\right)\right.\right.$$
The integral was offered to me by my good friend and it looks very difficult, I managed to solve only for $n=1$ and $n=2$.
$$\int_0^1 \frac{\arctan(x)\log^2(x)}{1+x} \, dx=\frac{21}{64}\pi\zeta(3)-\frac{\pi^3}{32}\log(2)-\frac{\pi^2}{24}G;$$
and
$$\int_0^1 \frac{\arctan(x)\log^4(x)}{1+x} \, dx=\frac{1395}{256}\pi\zeta(5)-\frac{9}{128}\pi^3\zeta(3)-\frac{7}{480}\pi^4G-\frac{5}{128}\pi^5\log(2)+\frac{\pi^6}{192}-\frac{\pi^2}{1536}\psi^{(3)}\left( \frac{1}{4} \right).$$
where $G$ represents the Catalan’s constant.
 A: The general "closed-form" is:
$$\int _0^1\frac{\arctan \left(x\right)\ln ^{2n}\left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{4}\left(1-2^{-2n}\right)\zeta \left(2n+1\right)\left(2n\right)!+\frac{1}{2}\beta \left(2n+2\right)\left(2n\right)!-\frac{\pi }{16}$$
$$\lim _{s\to 0}\left(\frac{d^{2n}}{ds^{2n}}\left(\csc \left(\frac{\pi s}{2}\right)\left(\psi \left(\frac{3-s}{4}\right)-\psi \left(\frac{1-s}{4}\right)\right)+\sec \left(\frac{\pi s}{2}\right)\left(\psi \left(1-\frac{s}{4}\right)-\psi \left(\frac{1}{2}-\frac{s}{4}\right)\right)-2\pi \csc \left(\pi s\right)\right)\right)$$
And you may find its evaluation in the book (Almost) Impossible Integrals, Sums, and Series through pages $140-145$ where double integration, symmetry and the following result are heavily exploited:
$$\int _0^{\infty }\frac{x^s}{\left(1+x\right)\left(1+y^2x^2\right)}\:dx=\frac{\pi }{2}\csc \left(\frac{\pi s}{2}\right)\frac{y^{-s}}{1+y^2}+\frac{\pi }{2}\sec \left(\frac{\pi s}{2}\right)\frac{y^{1-s}}{1+y^2}-\frac{\pi \csc \left(\pi s\right)}{1+y^2}$$
