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.

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.
