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So I've come across the following integral: $$\int _{\frac{-1}{\sqrt3}}^{\frac{1}{\sqrt3}}\frac{x^4}{1-x^4}\arccos\left(\frac{2x}{1+x^2}\right)\mathrm{d}x$$ And I have solved it like this: \begin{align} I&=\int_{\frac{-1}{\sqrt3}}^{\frac{1}{\sqrt3}}\frac{x^4}{1-x^4}\left(\frac{\pi}{2}-\arcsin\left(\frac{2x}{1+x^2}\right)\right)\mathrm{d}x\\[2ex] &=\frac{\pi}{2}\int_{\frac{-1}{\sqrt3}}^{\frac{1}{\sqrt3}}\frac{x^4}{1-x^4}\mathrm{d}x-\int_{\frac{-1}{\sqrt3}}^{\frac{1}{\sqrt3}}2\arctan x\cdot \frac{x^4}{1-x^4}\mathrm{d}x \\[2ex] &=\frac{\pi}{2}\int_{\frac{-1}{\sqrt3}}^{\frac{1}{\sqrt3}}\frac{x^4}{1-x^4}\mathrm{d}x\\[2ex] &=\pi\int_{0}^{\frac{1}{\sqrt3}}\frac{x^4}{1-x^4}\mathrm{d}x\\[2ex] &=\pi\int_{0}^{\frac{1}{\sqrt3}}\frac{x^4+1-1}{1-x^4}\mathrm{d}x\\[2ex] &=\pi\int_{0}^{\frac{1}{\sqrt3}}\frac{1}{1-x^4}-1\;\mathrm{d}x\\[2ex] &=\pi\int_{0}^{\frac{1}{\sqrt3}}\frac{1}{1-x^4} \mathrm{d}x\;-\frac{\pi}{\sqrt3}\\[2ex] &=\pi\int_{0}^{\frac{1}{\sqrt3}}\frac{1}{(1-x^2)(1+x^2)}\mathrm{d}x-\frac{\pi}{\sqrt3}\\[2ex] \text{substituting $x=\tan t:$}\\[2ex] &=\pi\int_0^{\frac{\pi}{6}}\frac{1}{1-\tan^2t}\mathrm{d}t-\frac{\pi}{\sqrt3}\\[2ex] &=\frac{\pi}{2}\int_0^{\frac{\pi}{6}}\frac{1+\cos2t}{\cos2t}\mathrm{d}t\\[2ex] &=\frac{\pi}{2}\left[\int_0^{\frac{\pi}{6}}\sec2t \; \mathrm{d}t+\int_0^{\frac{\pi}{6}}\mathrm{d}t\right]-\frac{\pi}{\sqrt3} \end{align} Applying the standard integral formulae and placing the limits, we end up with: $$\boxed{\frac{\pi^2}{12}+\frac{\pi}{4}\ln(2+\sqrt3)-\frac{\pi}{\sqrt3}}$$

Is there any other way to proceed with this integral ? Possibly without the trigonometric substitution?
Thanks.

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  • $\begingroup$ What happened to the 2 in the denominator of $\pi/2$ around the 4th line down? Did you mean to also change the lower integration limit to 0 perhaps? $\endgroup$ – user307169 May 28 at 15:41
  • $\begingroup$ When you go from the third line to the fourth line, $\frac\pi2$ changes to $\pi$ and nothing else changes. $\endgroup$ – saulspatz May 28 at 15:43
  • $\begingroup$ Yes. My mistake $\endgroup$ – sai-kartik May 28 at 15:44
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$\frac{1}{1-x⁴}\ = \ \frac{1+x²-(x²-1))}{2(1+x²)(1-x²)} \ =\ \frac{1}{2(1-x^2)}\ -\ \frac{1}{2(1+x^2)}$

These 2 are pretty standard forms

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$$\pi\int_{0}^{\frac{1}{\sqrt{3}}} \frac{\mathrm{d}x}{1-x^4}=\pi\int_{0}^{\frac{1}{\sqrt{3}}} \frac{1}{2}\left(\frac{1}{1+x^2}+\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)\right) \; \mathrm{d}x$$

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  • $\begingroup$ Ahh.. good ol' partial fraction $\endgroup$ – sai-kartik May 28 at 15:39
  • $\begingroup$ i haven't noticed this before, but you've made an error with the lower limit (just like I did). It should be $0$ $\endgroup$ – sai-kartik May 28 at 16:20
  • $\begingroup$ I just copied your integral up to the part where you were confused on and simplified it thereafter. I didn't check your preceding work. $\endgroup$ – Ty. May 28 at 16:35
  • $\begingroup$ I totally understand and accept it was my mistake. I too was copy-pasting and made the error :P. I have edited your answer though $\endgroup$ – sai-kartik May 28 at 16:41

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