closed form for $I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$ $$I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$$
for $n=1$ I tried to use $\arctan x=u$
and by notice that $$\frac{1+\tan u}{1-\tan u}=\cot\left ( \frac{\pi}{4}-u \right )$$ then $\frac{\pi}{4}-u=y$ and got
$$I(1)=\int_0^{\pi/4}y\cot y dy$$
which equal to $$I(1)=\frac{1}{8}(4G +\pi \log 2)$$
Where $G$ is Catalan constant

so $$I(n)=\int_0^{\pi/4}x^n\cot x dx$$
but how to find the closed form for any n using real or complex analysis ?
and what about $I(2)$ it seems related to $\zeta(3)$ ?!!

$I(2)={1\over64}\bigl(16G\pi+\pi^2\log(4)-35\zeta(3)\bigr)$, via Mathematica.  Not sure, however, of the derivation.  
