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.
 A: Hint:
Use the identity
$$1+x^4=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$$
and Partial fractions decomposition.
Edit:
Then
$$\dfrac{1}{1+x^4}=\dfrac{1}{(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)}\\=
\dfrac{Ax+B}{1+\sqrt{2}x+x^2}+\dfrac{Cx+D}{1+\sqrt{2}x+x^2}.$$
A: Note the following:


*

*$2 = (1 + x^{2}) + (1-x^{2})$.

*$\displaystyle \int \frac{1}{1+x^{4}} = \frac{1}{2} \int\frac{2}{1+x^{4}} = \frac{1}{2} \int\frac{(1+x^{2})+(1-x^{2})}{1+x^{4}} = \frac{1}{2} \int\frac{1+x^{2}}{1+x^{4}} + \frac{1}{2}\int\frac{1-x^{2}}{1+x^{4}} = \frac{1}{2} I_{1} + \frac{1}{2}I_{2}$.

*$\displaystyle I_{1} =\int\frac{1+x^{2}}{1+x^{4}} = \int\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} =\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2} +2} = \int\frac{1}{t^{2}+2}.$ 

*Note. We have made the substituion $t =x - \frac{1}{x}$. The integral $I_{2}$ can be evaluated in the same manner.
A: $$\frac{1}{1+x^4}=\frac{Ax+B}{2\sqrt2 (-x^2+\sqrt 2 x-1)}+\frac{Cx+D}{2\sqrt2 (x^2+\sqrt 2 x+1)}=\dots A=C=1, D=-B=\sqrt 2$$
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.
