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Evaluate $\int \frac{1}{1+x^4}dx$.

I am finding it very difficult to evaluate this integral. I do not know any standard formula for this integral. I cannot think of a suitable substitution. I tried to evaluate the integral by partial fractions but it become a bit messy and integration by parts does not result in a simpler integral. How can you evaluate this integral without using partial fractions?

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marked as duplicate by user228113, Jyrki Lahtonen Apr 9 '18 at 9:56

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  • $\begingroup$ Try using u = x^2 and the differentiation for tan inverse x $\endgroup$ – Brandon Loi Apr 9 '18 at 9:52
  • $\begingroup$ math.stackexchange.com/questions/426152/… $\endgroup$ – lab bhattacharjee Apr 9 '18 at 9:52
  • $\begingroup$ @BrandonLoi What you suggest won't work, because the integrand becomes, bar the mandatory considerations of sign, $\int^{x^2} \frac{1}{2(1+u^2)\sqrt u}\,du$ $\endgroup$ – user228113 Apr 9 '18 at 9:56
  • $\begingroup$ To all users. Approach0 is often better at finding duplicates than the on-site search engine (or Google) when it is essential to include a TeX-snippet. $\endgroup$ – Jyrki Lahtonen Apr 9 '18 at 9:59
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Use the fact that$$\frac1{x^4+1}=\frac1{x^4+2x^2+1-2x^2}=\frac1{\left(x^2+\sqrt2x+1\right)\left(x^2-\sqrt2x+1\right)}.$$

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Hint: $1+x^4=(1+\sqrt2x +x^2)(1-\sqrt2x +x^2)$.

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