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I understand it is an even function, which indicates $I=\frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{x^4+a^4}$

What should I do in the next step?

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    $\begingroup$ you can divide everything by $a^4$ to assume, essentially, $a=1$. Then you can either (1) use complex analysis or (2) factor the denominator into $(x^2-\sqrt{2}x-1)(x^2+\sqrt{2}x-1)$, use partial fractional decomposition, then integrate from there (using one or two other tricks such as completing the square). $\endgroup$ – mathworker21 Aug 18 '19 at 14:08
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    $\begingroup$ I'm sure you can find both steps done by googling "integral of 1/(x^4+1)" $\endgroup$ – mathworker21 Aug 18 '19 at 14:08
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Writing $$\frac{1}{a^4\left(\left(\frac{x}{a}\right)^4+1\right)}$$ now substitute $$t=\frac{x}{a}$$ so $$dt=\frac{1}{a}dx$$ and factorize $$t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-2t^2=(t^2+1-\sqrt{2}t)(t^2+1+\sqrt{2}t)$$

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    $\begingroup$ And how exactly does this gets the OP any closer to the solution? $\endgroup$ – mrtaurho Aug 18 '19 at 14:17
  • $\begingroup$ @mrtaurho By factorizing he can then split the expression in two fractions and continue form there... $\endgroup$ – Anastassis Kapetanakis Aug 18 '19 at 15:02
  • $\begingroup$ My comment was written before he added the line about factorization. Still I do not think that this is a good answer as it only gives a vague idea of how to evaluate the integral and additionally the comment of mathworker21 states precisely the same. $\endgroup$ – mrtaurho Aug 18 '19 at 15:04

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