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How to find the antiderivative of the function $f(x)=\frac{1}{1+x^4}$?

My math professor suggested to use the method of partial fractions, but it doesn't seem to work, because the denominator cannot be factored at all. Attempting to integrate with other familiar methods (integration by parts, trigonometric substitution) didn't work out as well. Does anyone have an idea what is the best approach to this problem?

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  • $\begingroup$ partial fraction decompostion: keep in mind that there are also imaginary solutions to the polynomial in the denominator $\endgroup$ – tired Jan 27 '16 at 1:22
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It's easy to see factorization in the first line. \begin{align*} \int \frac{1}{1+x^{4}}\mathrm{d}x &=\int \left ( \frac{\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}+\sqrt{2}x+1}+\frac{-\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}-\sqrt{2}x+1} \right )\mathrm{d}x\\ &=\int \left [ \frac{\dfrac{\sqrt{2}}{4}\left ( x+\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x+\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}}+\frac{-\dfrac{\sqrt{2}}{2}\left ( x-\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x-\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}} \right ]\mathrm{d}x \\ &=\frac{\sqrt{2}}{8}\ln\left | \frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1} \right |+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x+1 \right )+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x-1 \right )+C. \end{align*}

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It factors as the product of two quadratic polynomials. Irreducible polynomials in $\mathbf R[x]$ are linear polynomials and quadratic polynomials with negative discriminant.

Hint:

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

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