This is not a complete solution but it shows, for a simplified version of the problem, how a frequenty used approach works.
Consider the simplified problem to calculate the sum
$$s(k,x) = \sum _{n=0}^{\infty } \frac{P_n(x)}{k+n}$$
Where $k>0$ is a real parameter.
The generating function for the Legendre polynomials is
$$g(t,x) = \frac{1}{\sqrt{t^2-2 t x+1}} = \sum _{n=0}^{\infty } t^n P_n(x)$$
Now writing
$$\frac{1}{k+n}=\int_0^{\infty } \exp (-z (k+n)) \, dz$$
inserting this into the expression for $s$ we get
$$\sum _{n=0}^{\infty } P_n(x) \int_0^{\infty } \exp (-z (k+n)) \, dz$$
Interchanging summation and integration gives
$$\int_0^{\infty } \left(\sum _{n=0}^{\infty } P_n(x) \exp (-z (k+n))\right) \, dz$$
Extracting the factor $ \exp (- k z)$ from the sum this can be written as
$$\int_0^{\infty } \exp (- k z) \sum _{n=0}^{\infty } \exp (- n z) P_n(x) \, dz$$
Now the sum can be done using the formula for the generating function of the Legendre polynomials (with $t = \exp(- z))$ with the result
$$\sum _{n=0}^{\infty } \exp (-n z) P_n(x) = \frac{1}{\sqrt{-2 x e^{-z}+e^{-2 z}+1}}$$
Hence we arrive at this integral representation of $s$:
$$s1(k,x) = \int_0^{\infty } \frac{e^{-k z}}{\sqrt{-2 e^{-z} x+e^{-2 z}+1}} \, dz$$
Mathematica gives for this integral the closed form expression:
$$s1(k,x) = \frac{1}{k (k+1) (k+2)}\left( (k+1) (k+2) F_1\left(k;-\frac{1}{2},-\frac{1}{2};k+1;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)+2 k (k+2) x F_1\left(k+1;\frac{1}{2},\frac{1}{2};k+2;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)-k (k+1) F_1\left(k+2;\frac{1}{2},\frac{1}{2};k+3;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)\right)$$
Here $F_1$ is the Appell1 hypergeometric function (http://mathworld.wolfram.com/AppellHypergeometricFunction.html).
Admittedly, this is not a very "pleasant" expression, but the result for the complete problem of the OP can be expected to be even uglier, if it has a closed form at all, which I doubt.
EDIT
For integer $k$ we get closed form expressions (always for the simplified Problem). The first few are
$$s(0,x) = \log (2)-\log \left(-x+\sqrt{2-2 x}+1\right)$$
Notice that for $k=0$ the n-sum starts at $n=1$.
$$s(1,x) = \log \left(\frac{x-\sqrt{2-2 x}-1}{x-1}\right)$$
$$s(2,x) = \sqrt{2-2 x}+2 x \coth ^{-1}\left(\sqrt{2-2 x}+1\right)-1$$
$$s(3,x) = \frac{1}{2} \left(\left(6 x^2-2\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)$$
$$s(4,x) = \frac{1}{6} \left(6 x \left(5 x^2-3\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+5 x \left(3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)-2 \sqrt{2-2 x}+4\right)$$
Best Answer
The definition of $\eta(s)$ is $$\eta(s)=(1-2^{1-s})\zeta(s)$$ Take the derivative, and we get by product rule $$\eta'(s)=2^{1-s}\ln(2)\zeta(s)+(1-2^{1-s})\zeta'(s)$$ $\eta(s)$ also has the following series representation $$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$ Again, take the derivative $$\eta'(s)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^s}$$ Now, we can obtain a closed form $$\eta'(2)=\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}=2^{1-2}\ln(2)\zeta(2)+(1-2^{1-2})\zeta'(2)$$ Note that
$$\begin{align} \sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n^2}&=2^{-1}\ln(2)\zeta(2)+(1-2^{-1})\zeta'(2)\\ &=\frac{\pi^2}{12}\ln(2)+\frac{\pi^2}{12}(-12\ln(A)+\ln(2\pi)+\gamma)\\ &=\frac{\pi^2}{12}(\ln(4\pi)-12\ln(A)+\gamma) \end{align}$$