How do we integrate $\dfrac{1}{1+x^4}$? I am still confused with this integral: $\int \dfrac{1}{1+x^4}dx$
There one solution was given by a pupil that we write it as $\dfrac{x^2+1-(x^2-1)}{2(1+x^4)}$
Then split into two terms and proceed by putting $x+{1\over x}=t$ and completing the square.
My question is can we approach this question using other technique, most obviously partial fraction? The roots are complex, so how do we deal with it? Also please tell me a general procedure for such questions! Thank You!

For partial fractions factor into quadratics (this is the hard part):
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$
Do the standard decomposition method with undetermined cofficients:
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}
-\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right)$$
Integrate:
$$\int\frac{dx}{x^4+1}=
\frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)
+\frac{1}{2\sqrt{2}}((\arctan(\sqrt{2}x+1)+\arctan(\sqrt{2}x-1))$$
I think its quite beautiful.
