What is the correct approach for $\int\frac{x^{4}+1}{x^{6}+1}dx$? I was looking for tricky integrals to give something more challenging a try, and I stumbled upon [this] (Ignoring the definite part since I'm just interested in solving the integral):
$$\int\frac{x^{4}+1}{x^{6}+1}dx$$
My first reaction was to try substituting $[t=x^{2}; \frac{dx}{dt}=\frac{1}{2\sqrt{t}}]$, and everything went off the rails from there:
$$\int\frac{t^2}{t^3+1}\frac{1}{2\sqrt{t}}dt+\int\frac{1}{t^3+1}\frac{1}{2\sqrt{t}}dt$$
after that I tried getting $3t^2$ in the first integral, but it's pointless since it's a product and not an addition. I have also tried integration by parts, but I get things like $-\frac{2}{(2\sqrt{t})^3}$, which make everything way worse than it was before. There are no trigonometric identities involved, and I'm not sure I can apply rational integration since $x^6+1$ doesn't actually have any roots as far as I know. I have also tried other substitutions, like $[t=x^3]$, but I haven't been able to go further with those.
I'm totally out of ideas, I've checked all the books I have available for clues or methods I could have missed, but I didn't find a thing.
What am I missing? There is obviously an approach I have missed, I really don't think that substitution was the way to go. Any clues about what method to use? (I'm not looking for the solution)
 A: As $\displaystyle x^6+1=(x^2+1)(x^4-x^2+1)$
and $x^4-x^2+1=(x^2+1)^2-3x^2=(x^2+1-\sqrt3x)(x^2+1+\sqrt3x)$
Using 
Partial Fraction Decomposition,
we can write $$\frac{x^4+1}{x^6+1}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2-\sqrt3x+1}+\frac{Ex+F}{x^2+\sqrt3x+1}$$ where $A,B,C,D,E,F$ are arbitrary constants 
Now multiply either sides by $x^6+1$ and compare the coefficients of the different powers of $x$ to find $A,B,C,D,E,F$
Again as $x^2-\sqrt3x+1=\left(x-\frac{\sqrt3}2\right)^2+\left(\frac12\right)^2$
using Trigonometric substitution, set $x-\frac{\sqrt3}2=\frac12\tan\phi$
Similarly for $x^2+\sqrt3x+1$
A: An alternative, no need for partial fractions.
Starting the same way as lab bhattacharjee, note that $x^6 + 1 \equiv (x^2 + 1)(x^4 - x^2 + 1):$
$$\begin{align*} \int \frac{x^4 + 1}{x^6 + 1} dx = \int\frac{x^4 - x^2 + 1 + x^2}{(x^2 +1)(x^4 - x^2 + 1)}  dx & = \int\frac{x^4 - x^2 + 1}{(x^4 - x^2 + 1)(x^2 + 1)} dx + \int\frac{x^2}{x^6 + 1} dx \\ & = \int\frac{dx}{x^2 + 1} + \int\frac{x^2}{x^6 + 1} dx \\ & = \arctan x + \tfrac{1}{3} \arctan x^3 + \mathcal{C}\end{align*}$$
