Help with an Inverse Trigonometry Integral 2 Evaluate

$$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$


The Solution: We try to eliminate $\cos^{-1}\frac{2x}{1+x^2}$ by using the relation $$\pi - cos^{-1}(a) = cos^{-1}(a)$$
Consider the first 3 steps:$$$$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$
Putting $x=-t$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{(-t)^4}{1-(-t)^4}\cos^{-1}\frac{2(-t)}{1+(-t)^2} (-dt)$$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t) $$ $$$$
Now, how do we add both of these:
$$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x +\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t)$$ Could somebody please show this in detail?
 A: Let $I$ be given by 
$$I=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx$$
Upon substituting $x\to -x$ we find that 
$$\begin{align}
I&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{-2x}{1+x^2}\right)dx\\\\
&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \left(\pi-\arccos\left(\frac{2x}{1+x^2}\right)\right)dx\\\\
&=\pi\,\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4}dx-\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx\\\\
&=\pi\,\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4}dx-I
\end{align}$$
Thus, 

$$I=\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} dx$$


The integral $I=\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} dx$ can be evaluated in closed form.  First we note that
$$\frac{x^4}{1-x^4}=-1+\frac{1}{1-x^4}$$
Thus, $I$ becomes
$$I=-\pi\sqrt{3}/3+\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1-x^4} dx$$
Next, we use partial fraction expansion to write 
$$\frac{1}{1-x^4}=\frac{1/4}{1-x}+\frac{1/4}{1+x}+\frac{1/2}{1+x^2}$$
whereby we find that 
$$\begin{align}
I&=-\pi\sqrt{3}/3+\frac{\pi}{8} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1-x}dx+\frac{\pi}{8} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1+x}dx+\frac{\pi}{4} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1+x^2}dx\\\\
&=-\pi\sqrt{3}/3+\frac{\pi}{4}\log\left(2+\sqrt{3}\right)+\pi^2/12\\\\
&=\frac{\pi}{4}\log(2+\sqrt{3})+\frac{\pi^2}{12}-\frac{\pi}{\sqrt{3}}
\end{align}$$
Finally, $abcdef+1=577$.
