Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:

$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$

I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also am unable to think substitution.

$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\sin^4 x + \cos ^4 x =\frac{1}{4}\left[(1- \cos2x)^2 +(1+ \cos2x)^2 \right]
=\frac{1}{2}((1+ \cos^2 2x)$$
$$\implies 2 \int \frac{dx}{1+ \cos^2 2x}$$
$$= 2 \int \frac{\sec^2 2x}{\sec^2 2x+ 1}{dx}$$
$$t= \tan2x$$
$$\implies 2 \int \frac{dt}{2 + t^2}$$
$$\frac{1}{\sqrt2} \tan^{-1} \left(\frac{t}{\sqrt2}\right)$$
