Evaluate the following indefinite integral. $\int\frac{1}{x^4+1}\, dx$

Question

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

Answer 1

Hint:

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

Answer 2

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.