Evaluate the integral:
$$\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\left(\frac{x^4}{1-x^4}\right) \arccos \left(\frac{2x}{1+x^2}\right)\, {\rm d}x$$
Solution
We have successively:
\begin{align*}\mathcal{J}=\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\arccos \left ( \frac{2x}{1+x^2} \right )\,{\rm d}x &= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\arccos \left ( -\frac{2x}{1+x^4} \right )\,{\rm d}x\\&= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\left [ \pi-\arccos \left ( \frac{2x}{1+x^4} \right ) \right ]\,{\rm d}x\\&=\pi \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d} x-\mathcal{J} \end{align*}
Thus:
$$\mathcal{J}=\frac{\pi}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d}x \implies \mathcal{J}=-\frac{2}{\sqrt{3}}+\frac{\pi}{6}+\frac{1}{2}\ln \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )$$
$$\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\left(\frac{x^4}{1-x^4}\right) \arccos \left(\frac{2x}{1+x^2}\right)\, {\rm d}x$$
Solution
We have successively:
\begin{align*}\mathcal{J}=\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\arccos \left ( \frac{2x}{1+x^2} \right )\,{\rm d}x &= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\arccos \left ( -\frac{2x}{1+x^4} \right )\,{\rm d}x\\&= \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\left [ \pi-\arccos \left ( \frac{2x}{1+x^4} \right ) \right ]\,{\rm d}x\\&=\pi \int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d} x-\mathcal{J} \end{align*}
Thus:
$$\mathcal{J}=\frac{\pi}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{x^4}{1-x^4}\,{\rm d}x \implies \mathcal{J}=-\frac{2}{\sqrt{3}}+\frac{\pi}{6}+\frac{1}{2}\ln \left ( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right )$$
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