<s>[INST] how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.

I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,

How I can use this to solve of that integration.
 [/INST] $$\frac{1}{1+x^4}=\frac{Ax+B}{2\sqrt2 (-x^2+\sqrt 2 x-1)}+\frac{Cx+D}{2\sqrt2 (x^2+\sqrt 2 x+1)}=\dots A=C=1, D=-B=\sqrt 2$$
Simplify even further 
$$\frac{x-\sqrt 2}{2\sqrt2 (-x^2+\sqrt 2 x-1)}= -\frac{\sqrt 2-2x}{2-x^2+\sqrt 2 x-1)}-\frac{1}{\sqrt 2(-x^2+\sqrt 2 x-1}$$
Substitute $u=-x^2+\sqrt 2 x-1$ then it's trivial. The other is quite similar.
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