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

We are asked to find the following anti-derivative $$\int \frac{dx}{1+x^4}$$

I tried partial fraction decomposition but I did not manage. How can I obtain the solution for the integral above?

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marked as duplicate by Hans Lundmark, Old John, Davide Giraudo, Paul, Matt Pressland Dec 19 '12 at 13:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers

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The steps are as follows:
1) Decompose $\frac{1}{1+x^{4}}$ using partial fractions (it can be factored using an identity of Sophie Germain)
2) You should have a linear function in each numerator and a quadratic in each denominator. Separate into the form $\frac{const}{quadratic}+\frac{const\cdot x}{quadratic}$
3) Complete the square on this quadratic.
4) To integrate the first form, make a simple substitution to transform the integrand into the form $\frac{1}{1+u^{2}}$, which is the derivative of $\tan^{-1}(x)$.
5) For the second, make another substitution to transform the integrand into the form $\frac{1}{1+v}$, which has antiderivative $\ln(1+v)$.

Be very careful with tiny algebraic slips, and keep track of your constants.

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Thank you! I managed to integrate following your steps. It is interesting how complicated the integration of such a simple integral is in the end. –  nesseril Dec 19 '12 at 12:31
Indeed! I must say, though, the top answer in Integration of $\int\frac{1}{1+x^{4}}dx$ is far nicer than mine. –  Daniel Littlewood Dec 19 '12 at 13:28
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HINT (for partial fractions) : $$ x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1). $$

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It is interesting that $\sqrt{2}$ appears suddenly. I did not manage to separate it into factors. –  nesseril Dec 19 '12 at 12:30
$x^4+1$ has 4 complex roots (the 4th roots of $-1$) which come in pairs $w,\bar w$ and $z, \bar z$. Then one sees that $w+\bar w=\pm\sqrt{2}$ and $z+\bar z=\mp\sqrt 2$. –  Andrea Mori Dec 19 '12 at 12:42
On a more elementary tune, consider the intermediate step $x^4+1=(x^2+1)^2-2x^2$. Here the appearence of $\sqrt2$ is immediately evident. –  Andrea Mori Dec 19 '12 at 12:44
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