Not an answer. Too long for comment and might distract if added in the question as edit.
I tried to get a closed form of $\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)]$
$$\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)] = -4 \sum\limits_{k=1}^{N} \int\limits_{0}^{k} \tan^{-1}(2x) dx = I$$
$$I = -4 \int\limits_{0}^{\infty} \tan^{-1}(2x) S_{N}(x)$$
Here, $S_{N}(x)$ is a staircase defined as
$$
S_{N}(x) =
\begin{cases}
N, & 0\le x \lt 1 \\
N-1, & 1 \le x \lt 2 \\
N-2, & 2 \le x \lt 3 \\
... \\
1, & N-1 \le x \lt N \\
0, & \text{elsewhere} \\
\end{cases}
$$
Then I created an odd-symmetry signal $M_{N}$ by flipping $S_{N}$ horizontally around y-axis and vertically on x-axis and adding to $S_{N}$.
$$M_{N}(x) = S_{N}(x)-S_{N}(-x)$$
So, $$I=-2\int\limits_{-\infty}^{\infty} \tan^{-1}(2x) M_{N}(x)$$
Now, let us use Fourier Transforms and then use Parseval Identity.
Fourier Transform formula I use is $FT[f(x)](y) = \int\limits_{-\infty}^{\infty} f(x) e^{-i2 \pi x y}dx$
With this, $$FT[\tan^{-1}(2x)](y) = -i \frac{e^{-|\pi y|}}{2y}$$ and
$$FT[M_{N}(x)](y) = \frac{i}{2} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1]$$
Using Parseval Identity ($\color{red}{\mathbf{\text{Edit: Plancherel/Parseval identity is}}}$ $\int\limits_{-\infty}^{\infty} f(x) \overline{g(x)} dx = \int\limits_{-\infty}^{\infty} \hat{f}(\xi) \overline{\hat{g}(\xi)} d\xi$. I had overlooked the conjugate and hence removing the negative sign I had earlier.),
$$I = \frac{1}{2}\int\limits_{-\infty}^{\infty} \frac{e^{-|\pi y|}}{y} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1] dy$$
Because the integrand is even,
$$I = \int\limits_{0}^{\infty} \frac{e^{-|\pi y|}}{y} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1] dy$$
Substituting $\pi y = t$,
$$I = \int\limits_{0}^{\infty} \frac{e^{-t}}{t} \operatorname{sinc}(t) \operatorname{csc}(t)[\operatorname{csc}(t) \sin\{(2N+1) t\} -2N-1] dt$$
Expand $\operatorname{sinc}(t)$ as $\frac{\sin(t)}{t}$ to get rid of one $\operatorname{csc}(t)$.
$$I = \int\limits_{0}^{\infty} \frac{e^{-t}}{t^2} [\operatorname{csc}(t) \sin\{(2N+1) t\} -2N-1] dt$$
Note: I have verified that this integral is accurate by comparing it for small values of $N$.
I haven't proceeded much afer this. Mathematica cannot give me a closed form of this (which I know will diverge when $N \to \infty$). The hope is to get a closed form from this without summations (and most importantly not bring back the Polylog I am solving) and then do a series expansion at $N = \infty$ and plug it back into my original question in the place of my summation that I currently have. I again hope that such a series will be in terms of logarithms that might hopefully be amenable to a limit calculations. Enough of my hopes. I request all the integration masters on this site to help solving this integral if you find the above steps useful, or else, give another closed form for my original summation $\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)]$