Evaluate $\int_{0}^{\infty} \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}\>dx$ How to Integrate
$$  I = \int_{0}^{\infty} \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \>dx\approx -0.295512 $$
Mathematica returns a result that does not match numerically with the integral approximation :
$$ \frac{i C}{4} -\frac{\pi}{4}\sqrt{4-3i} +\frac{3 \pi^2}{32} +\pi\left(\frac{1}{4}-\frac{i}{8}\right)\coth^{-1}(\sqrt{2}) \approx -0.048576 + 0.43823\,i $$

Where C denotes Catalan's Constant
Motivation
I was able to find a closed form for
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
using double infinite sums. Upon plotting the function within the integral i saw that it could be integrated from $0$ to ${\infty}$ .
Attempts
Number 1
I tried to Split the integral as
$$ \int_{0}^{1} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx + \int_{1}^{\infty} \frac{\tan^{-1}(x^2)}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)}dx $$
and use the Taylor Series for $\tan^{-1}(x^2) $ when $|x| >1$
Number 2
I tried using partial fractions as
$$ \frac{1}{x^4-1} = \frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^2+1)} $$
$$ \frac{1}{x^2(x^4-1)} = \frac{1}{2(x^2+1)}-\frac{1}{x^2}-\frac{1}{4(x+1)} + \frac{1}{4(x-1)} $$
from which I obtained
$$\int_{0}^{\infty} \frac{\tan^{-1}(x^2)}{2(x^2+1)}dx = \frac{\pi^2}{16} $$
$$ \int_{0}^{\infty} \frac{\pi}{8(x^2+1)} dx= \frac{\pi^2}{8} $$
but was unable to proceed further.
Number 3
A number of basic integration techniques such as U-Sub and Integration by parts.
Number 4
Using the same technique i used to evaluate the same integral but from $0$ to $1$
I will continue to try , but for now I find myself to be stuck.
Q - Is there a closed form for I? If the solution is easy and i am missing something , could you provide hints instead?
Thank you for your help and time.
 A: Note
$$\frac1{x^4-1} =\frac1{2(x^2-1)} - \frac1{2(x^2+1)}\\
\frac1{x^2(x^4-1)} =\frac1{2(x^2-1)} + \frac1{2(x^2+1)}-\frac1{x^2}
$$
and rewrite the integral as
\begin{align}
 I &= \int_{0}^{\infty} \left( \frac{\tan^{-1}x^2}{x^2(x^4-1)}-\frac{\pi}{4(x^4-1)} \right)dx\\
&=\frac{\pi^2}{16}+ \frac12\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx
 - \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx
 + \frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\tag1
\end{align}
where $\int_0^{\infty} \frac{\tan^{-1}x^2}{1+x^2}dx=\frac{\pi^2}8$, $
 \int_0^{\infty} \frac{\tan^{-1}x^2}{x^2}dx=\frac{\pi}{\sqrt2}$ and
\begin{align}
&\frac12\int_0^{\infty} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx
=\int_0^{1} \frac{\tan^{-1}x^2-\frac\pi4}{x^2-1}dx\\
\overset{IBP}=&-\int_0^1 \frac x{1+x^4} \ln \frac{1-x}{1+x}dx
\overset{x\to \frac{1-x}{1+x}}=-\int_0^\infty\frac {\ln x}{1+6x^2+ x^4}dx\\
 =&-\frac1{4\sqrt2}\left( \int_0^\infty\frac{\ln x}{x^2+ (\sqrt2-1)^2}dx -\int_0^\infty\frac{\ln x}{x^2+ (\sqrt2+1)^2}dx \right)\\
 =&-\frac1{8\sqrt2}\left(\frac{\ln(\sqrt2-1)}{\sqrt2-1} - \frac{\ln(\sqrt2+1)}{\sqrt2+1} \right)
\end{align}
Substitute above results into (1) to obtain
$$I= \frac{\pi^2}8-\frac\pi{\sqrt2}\left(1+\frac{\ln(\sqrt2-1)}{8(\sqrt2-1)} - \frac{\ln(\sqrt2+1)}{8(\sqrt2+1)} \right)
$$
A: I haven't tested this myself, but this problem reminds me of something I saw in complex analysis with the Residue Theorem. For the second part (after the minus sign) a good contour would just be the upper half of a semicircle, in combination with a line along the Real axis (see picture). Again, I have not yet tested if the correct integrals cancel out, but this is my current guess for an additional approach.

For the first part of the integral, a contour like below may be more appropriate, since 0 is also a singularity.

The reason we like these particular contours is that we should somehow show the complex part of the path is 0 (or apply Jordan's Lemma, if we can), and so we can limit the Real axis piece to get $[-\infty, \infty]$ and get the Cauchy principal value. Further, your functions are even functions, so we can then at the end split in half to get the upper half of the integral.
This is a guess! I will attempt and edit if there are troubles, or if I think of another way!
EDIT:
Because of singularities along the real axis, you will have to do something closer to the second contour, with the indent, around $z = \pm 1$. But writing it all down, I believe this integral is not too difficult (i.e., can be done by hand) using this method of contour integration, and then evenness at the end.
