How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$? 
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$ ?  

The numerator is a irreducible polynomial so I can't use partial fractions.
I tried the substitutions $t = x^2, t=x^4$ and for the formula $\int u\,dv = uv - \int v\,du$ I tried using: $u=\frac{x^4 + 1 }{x^6 + 1} , \,dv=\,dx \\ u=\frac{1}{x^6 + 1} , \,dv= (x^4 + 1) \,dx \\u=x^4 + 1 , \,dv=\frac{\,dx}{x^6 + 1}$  
But I always get more complicated integrals.
Any hints are appreciated! 

$$\int \frac{x^4 + 1 }{x^6 + 1}=\int \frac{x^4 -x^2+ 1 }{x^6 + 1} + \int \frac{x^2 }{x^6 + 1}=\int \frac{1}{x^2 + 1} + \int \frac{x^2 }{x^6 + 1}$$
The first integral is known while the second is easy by $u=x^3$.
