Evaluate $\int_0^1\left(\frac{x-1}{x+1}\right)^n\frac{1}{\ln x} \,dx$ 
Evaluate: $$ I(n) = \int_0^1\left(\frac{x-1}{x+1}\right)^n\frac{1}{\ln x} \,dx$$
For $n \in \mathbb N$.

I started by noticing that $\int_0^1 x^y \,dy = (x-1)/\ln x$
So I can convert this into a double integral as follows:
$$I(n) = \int_0^1\int_0^1\frac{(x-1)^{n-1}x^y}{(x+1)^n}\,dy\,dx$$
Changing the order of integration and plugging in $x \rightarrow \cos 2\theta$, I get:
$$I(n) = 2(-1)^n\int_0^1 \int_0^{\pi/4}\tan^{2n-1}\theta \cos^y 2\theta \,d\theta \,dy$$
I have a feeling that I can somehow transform the inner integral into a Beta function, but the $\pi/4$ and $2\theta$ are a bit irritating and hard to get rid of.
Any methods to tackle this?
 A: Write $I_n$ for the integral and we introduce the following regularization:
$$ I_n(s) := - \int_{0}^{1} \left(\frac{x-1}{x+1}\right)^n (-\log x)^{s-1} \, \mathrm{d}x. $$
This defines an analytic function for $\operatorname{Re}(s) > -n$. We aim at determining the expression of $I_n(s)$ using the principle of analytic continuation.
To this end, we temporarily assume that $s > n$. Then applying the substitution $x \mapsto e^{-x}$,
\begin{align*}
I_n(s)
&= - \int_{0}^{\infty}\left(2 - \frac{1}{1+e^{-x}} \right)^n x^{s-1}e^{-x} \, \mathrm{d}x \\
&= - \sum_{k=0}^{n} \binom{n}{k} (-2)^k \int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{(1+e^{-x})^k} \, \mathrm{d}x
\end{align*}
Now we utilize the following expansion
$$ \frac{z}{(1+z)^k} = \frac{1}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \sum_{l=1}^{\infty} (-1)^{l-1} l^j z^l, $$
valid for $k \geq 1$ and $|z| < 1$, where $\left[{n \atop k}\right]$ is the unsigned Stirling numbers of the first kind. Then for $k \geq 1$, using the fact that $s > n$, Fubini's Theorem yields
\begin{align*}
\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{(1+e^{-x})^k} \, \mathrm{d}x
&= 
\frac{1}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \sum_{l=1}^{\infty} (-1)^{l-1} l^j \int_{0}^{\infty} x^{s-1}e^{-lx} \, \mathrm{d}x \\
&= 
\frac{1}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \sum_{l=1}^{\infty} (-1)^{l-1} \frac{\Gamma(s)}{l^{s-j}} \\
&= 
\frac{1}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \Gamma(s)\eta(s-j),
\end{align*}
where
$$ \eta(s) := \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = (1 - 2^{1-s})\zeta(s) $$
is the Dirichlet eta function. Plugging this back, we get
\begin{align*}
I_n(s)
&= - \Gamma(s) \Biggl( 1 + \sum_{k=1}^{n} \binom{n}{k} \frac{(-2)^k}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \eta(s-j) \Biggr).
\end{align*}
Although this equality is initially derived only for $s > n$, both sides define meromorphic functions on the region $\operatorname{Re}(s) > -n$, and as such, they must coincide on all of this region. Then, taking the limit as $s\to0$,
\begin{align*}
I_n = I_n(0)
&= - \sum_{k=1}^{n} \binom{n}{k} \frac{(-2)^k}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \eta'(-j) \\
&= - \sum_{k=1}^{n} \binom{n}{k} \frac{(-2)^k}{(k-1)!} \sum_{j=0}^{k-1} \left[ {k-1 \atop j} \right] \left( 2^{1+j} \zeta(-j)\log 2 + (1-2^{1+j})\zeta'(-j) \right).
\end{align*}
The following table is generated by Mathematica 11 using the above formula:
$$ \begin{array}{c|c}
\hline
n & I_n \\
\hline
 1 & \log (2 \pi )-2 \log (2) \\
 2 & -12 \log (A)+1-\frac{8 \log (2)}{3}+2 \log (2 \pi ) \\
 3 & -24 \log (A)-28 \zeta '(-2)+2-\frac{10 \log (2)}{3}+3 \log (2 \pi ) \\
 4 & -40 \log (A)+40 \zeta '(-3)-56 \zeta '(-2)+\frac{10}{3}-\frac{176 \log (2)}{45}+4 \log (2 \pi ) \\
 5 & -56 \log (A)-\frac{124}{3} \zeta '(-4)+80 \zeta '(-3)-\frac{308}{3} \zeta '(-2)+\frac{14}{3}-\frac{202 \log (2)}{45}+5 \log (2 \pi ) \\
\hline
\end{array} $$
