Integration Mania Continued

@Vincent Moroney 's solution. First we utilize the following integral property $\int_{-a}^a f(x)\,dx = \int_0^{a} f(x)\,dx + \int_0^{a} f(-x)\,dx$ and the property that $\arccos(-x) = \pi - \arccos(x)$ to give $\int_0^{\frac{1}{\sqrt{3}}} \frac{x^4}{1-x^4}\Big[\arccos\Big(\frac{2x}{1+x^2} \Big) + \arccos\Big(\frac{-2x}{1+x^2} \Big) \Big] \,dx$ $= \pi \int_0^{\frac{1}{\sqrt{3}}}\frac{x^4}{1-x^4} \,dx = - \pi \int_0^{\frac{1}{\sqrt{3}}} \frac{x^4}{x^4-1} \,dx.$ The remaining integral is easily computed with partial fraction decomposition: $\frac{x^4}{x^4-1} = 1 + \frac{1}{x^4-1} \Rightarrow \frac{1}{x^4-1} = \frac{1}{4}\frac{1}{x-1} -\frac{1}{4}\frac{1}{x+1} -\frac{1}{2}\frac{1}{x^2+1}$ so all that remains is the computation of the following integrals $-\pi \int_0^{\frac{1}{\sqrt{3}}} \Big(1 + \frac{1}{4}\frac{1}{x-1} -\frac{1}{4}\frac{1}{x+1} -\frac{1}{2}\frac{1}{x^2+1}\Big) \,dx .$ Since these are all trivial integrals, computation will be skipped. The desired answer is: $-\pi\Big(\frac{1}{\sqrt{3}}- \frac{\pi}{12} -\frac{1}{4} \ln \Big| \frac{\sqrt{3} +1}{\sqrt{3}-1} \Big|\Big)$ so $A+B+C+D+E+F+G+H = \boxed{26}$