The closed form representations of Integrals of logarithm functions I wish to find a closed form representations of the following integral
$$\int\limits_{0}^1\frac{\log^p(x)\log^r\left(\frac{1-x}{1+x}\right)}{x}dx=?$$
Here $p\ge 1$ and $r\ge 0$ are nonnegative integers. It can be expressed in terms of a linear combination of well known constants (such as: Riemann zeta values,$\pi$ et. al.)? 

Here is an alternative approach for the case $r = 1, p \in \mathbb{N}$ that has already been given by @Ininterrompue.
Let
$$I_p = \int_0^1 \ln^p x \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{x}.$$
Since
$$\ln \left (\frac{1 - x}{1 + x} \right ) = - 2 \sum_{n = 0}^\infty \frac{x^{2n+1}}{2n + 1}, \qquad |x| < 1,$$
the integral may be written as
$$I_p = -2 \sum_{n = 0}^\infty \frac{1}{2n + 1} \int_0^1 x^{2n} \ln^p x \, dx.$$
Integrating by parts $p$-times gives
\begin{align}
I_p &= - 2 (-1)^p p! \sum_{n = 0}^\infty \frac{1}{(2n + 1)^{p + 2}}\\
&= -\frac{(-1)^p p!}{2^{p + 1}} \sum_{n = 0}^\infty \frac{1}{(n + 1/2)^{p + 2}}\\
&= -\frac{(-1)^p p!}{2^{p + 1}} \cdot \frac{(-1)^{p + 2}}{(p + 1)!} \psi^{(p + 1)} \left (\frac{1}{2} \right )\\
&= -\frac{1}{(p + 1) 2^{p + 2}} \psi^{(p + 1)} \left (\frac{1}{2} \right ). \tag1
\end{align}
Here $\psi^{(m)} (x)$ is the polygamma function with the following series representation of
$$\psi^{(m)} (z) = (-1)^{m + 1} m! \sum_{n = 0}^\infty \frac{1}{(n + z)^{m + 1}},$$
having been used. 
Finally, as (see here)
$$\psi^{(m)} \left (\frac{1}{2} \right ) = (-1)^{m + 1} m! (2^{m + 1} - 1) \zeta (m + 1),$$
in terms of the  Riemann zeta function, $\zeta (z)$, one can rewrite (1) as
$$I_p = (-1)^{p + 1} p! \left (1 - \frac{1}{2^{p + 2}} \right ) \zeta (p + 2).$$ 
