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This question already has an answer here:

i want find the answer and metod of solve of $\int\frac{1}{1+x^4}dx$.

I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,

How I can use this to solve of that integration.

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marked as duplicate by abiessu, Norbert, Hagen von Eitzen, Vladimir Reshetnikov, Mark Bennet Oct 14 '14 at 22:32

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  • $\begingroup$ There's a way through complex analysis, but you need en.wikipedia.org/wiki/Residue_theorem $\endgroup$ – Exodd Oct 14 '14 at 19:34
  • $\begingroup$ @Exodd You could use complex analysis to find $\int_{-\infty}^\infty \frac1{1+x^4}\,dx$, but not the indefinite integral. $\endgroup$ – Mike Earnest Oct 14 '14 at 19:38
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    $\begingroup$ @abiessu, this question asks for an anti-derivative, not a definite integral. The OP might enjoy reading the first part of Git Gud's answer at math.stackexchange.com/questions/446902/… $\endgroup$ – Barry Cipra Oct 14 '14 at 19:50
  • $\begingroup$ @BarryCipra: I noticed that just after I clicked close. It looks like no one else thinks it was a duplicate, so it will just be a linked question then... $\endgroup$ – abiessu Oct 14 '14 at 19:58
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Hint:
Use the identity $$1+x^4=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$$ and Partial fractions decomposition.
Edit:
Then $$\dfrac{1}{1+x^4}=\dfrac{1}{(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)}\\= \dfrac{Ax+B}{1+\sqrt{2}x+x^2}+\dfrac{Cx+D}{1+\sqrt{2}x+x^2}.$$

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    $\begingroup$ how can use this? @M. Strochyk $\endgroup$ – DnShVr Oct 14 '14 at 19:44
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$$\frac{1}{1+x^4}=\frac{Ax+B}{2\sqrt2 (-x^2+\sqrt 2 x-1)}+\frac{Cx+D}{2\sqrt2 (x^2+\sqrt 2 x+1)}=\dots A=C=1, D=-B=\sqrt 2$$

Simplify even further

$$\frac{x-\sqrt 2}{2\sqrt2 (-x^2+\sqrt 2 x-1)}= -\frac{\sqrt 2-2x}{2-x^2+\sqrt 2 x-1)}-\frac{1}{\sqrt 2(-x^2+\sqrt 2 x-1}$$

Substitute $u=-x^2+\sqrt 2 x-1$ then it's trivial. The other is quite similar.

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Note the following:

  • $2 = (1 + x^{2}) + (1-x^{2})$.

  • $\displaystyle \int \frac{1}{1+x^{4}} = \frac{1}{2} \int\frac{2}{1+x^{4}} = \frac{1}{2} \int\frac{(1+x^{2})+(1-x^{2})}{1+x^{4}} = \frac{1}{2} \int\frac{1+x^{2}}{1+x^{4}} + \frac{1}{2}\int\frac{1-x^{2}}{1+x^{4}} = \frac{1}{2} I_{1} + \frac{1}{2}I_{2}$.

  • $\displaystyle I_{1} =\int\frac{1+x^{2}}{1+x^{4}} = \int\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} =\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2} +2} = \int\frac{1}{t^{2}+2}.$

  • Note. We have made the substituion $t =x - \frac{1}{x}$. The integral $I_{2}$ can be evaluated in the same manner.

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