Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2}x-1)}+\frac{x+\sqrt{2}}{2\sqrt{2}(x^2+\sqrt{2}x+1)}$$
BTW, are there any easier method to solve this integral? As seen on the answer, it seems not very possible. ;)

I am giving a hint.
$$\int \frac {dx}{1+x^4}=\frac 12\left(\int\frac {1+x^2}{1+x^4}dx-\int\frac {x^2-1}{1+x^4}dx\right)\\=\frac 12\left(\int\frac {1+\frac 1{x^2}}{x^2+\frac 1{x^2}}dx-\int\frac {1-\frac 1{x^2}}{x^2+\frac 1{x^2}}dx\right)$$
Now, write $x^2+\frac 1{x^2}=\left(x-\frac 1x\right)^2+2$ and $x^2+\frac 1{x^2}=\left(x+\frac 1x\right)^2-2$, for the respective integrals, substitute for $x+\frac 1x=z$ and $x-\frac 1x=u$.
Now, you have $\left(1-\frac 1{x^2}\right)dx=dz$ and $\left(1+\frac 1{x^2}\right)dx=du$
Now,proceed from here, it is easy for you now.
