How to show the following

$$\displaystyle \int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}dx \\ = \pi\sqrt{\frac{1+\sqrt{13}}{78}} \log \left(\frac{1+\sqrt{13}+\sqrt{2 \sqrt{13}-2}}{4} \right)+\pi\sqrt{\frac{\sqrt{13}-1}{78}}\tan ^{-1}\left(\frac{\sqrt{5+2 \sqrt{13}}}{3} \right)$$

It seems that it can be transformed into an elliptic integral. Maybe we can define

$$I(a) = \int_{0}^{1}\dfrac{\arcsin{a\sqrt{x}}}{x^4-2x^3+2x^2-x+1}dx $$

Then by differentiation

$$I'(a) = \int_{0}^{1}\dfrac{\sqrt{x}}{(\sqrt{1-a^2x})(x^4-2x^3+2x^2-x+1)}dx $$

Another approach might involve series expansion of $\arcsin(x)$

$$\arcsin(x) = \sum_{n\geq 0}\frac{{2n \choose n} }{4^n(2n+1)}x^{2n+1}$$

To be honest it seems helpless.




marked as duplicate by Zaid Alyafeai, Community Apr 26 '17 at 9:57

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