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I would like to evaluate the following indefinite integral:
$\int\frac{1}{x^4+1}\, dx$

user356595
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  • Use Sophie Germain's identity https://www.artofproblemsolving.com/wiki/index.php?title=Sophie_Germain_Identity, followed by a slightly tedious excursion into partial fractions. – David Quinn Aug 09 '16 at 18:44
  • @DavidQuinn Weirdly, that link appears to be broken for me due to some %E2%80%8C%E2%80%8B at the end of it, whereas https://www.artofproblemsolving.com/wiki/index.php?title=Sophie_Germain_Identity works fine for me. – Caleb Stanford Aug 09 '16 at 19:58

2 Answers2

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You may exploit: $$ x^4+1 = (x^2+1)^2-2x^2 = (x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1) \tag{1}$$ to write $\frac{1}{1+x^4}$ as $$ \frac{1}{1+x^4} = \frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+x\sqrt{2}+1}-\frac{x-\sqrt{2}}{x^2-x\sqrt{2}+1}\right).\tag{2}$$ After that, $$ \int\frac{x\pm\sqrt{2}}{x^2\pm x\sqrt{2}+1}\,dx = C+\frac{1}{2}\log\left(x^2\pm x\sqrt{2}+1\right)+\arctan\left(1\pm x\sqrt{2}\right)\tag{3} $$ settles the question.

Jack D'Aurizio
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  • I am not sure about the signs implied by $+\dfrac{1}{2}\log\left(x^2\pm x\sqrt{2}+1\right)$ when your previous line includes $-\dfrac{x-\sqrt{2}}{x^2-x\sqrt{2}+1}$ – Henry Aug 09 '16 at 18:58
  • Signs apart, the essence is that we may compute the primitive of both terms appearing in the RHS of $(2)$. – Jack D'Aurizio Aug 09 '16 at 19:02
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Hint:

$x^4+1 = (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$

Henry
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