TITLE: What is the correct approach for $\int\frac{x^{4}+1}{x^{6}+1}dx$?
QUESTION [6 upvotes]: 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)

REPLY [13 votes]: An alternative, no need for partial fractions.
Starting the same way as lab bhattacharjee, note that $x^6 + 1 \equiv (x^2 + 1)(x^4 - x^2 + 1):$
$$\begin{align*} \int \frac{x^4 + 1}{x^6 + 1} dx = \int\frac{x^4 - x^2 + 1 + x^2}{(x^2 +1)(x^4 - x^2 + 1)}  dx & = \int\frac{x^4 - x^2 + 1}{(x^4 - x^2 + 1)(x^2 + 1)} dx + \int\frac{x^2}{x^6 + 1} dx \\ & = \int\frac{dx}{x^2 + 1} + \int\frac{x^2}{x^6 + 1} dx \\ & = \arctan x + \tfrac{1}{3} \arctan x^3 + \mathcal{C}\end{align*}$$