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):


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:


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)

  • 1
    $\begingroup$ Introduce $u = x-x^{-1}$, $$ \int \frac{1+ x^4}{1+x^6} dx = \int \frac{x^2 + x^{-2}}{x^3 + x^{-3}}\frac{dx}{x} = \int \frac{x^2 + x^{-2}}{x^3 + x^{-3}}\frac{d(x-x^{-1})}{x+x^{-1}} = \int\frac{(u^2+2)du}{(u^2+4)(u^2+1)}\\ = \frac13 \int \left(\frac{1}{1+u^2}+\frac{2}{4+u^2}\right) du = \frac13 \left(\tan^{-1}u + \tan^{-1}\frac{u}{2}\right) + \text{const.} $$ $\endgroup$ – achille hui Feb 1 '14 at 17:53

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 $x^2+\sqrt3x+1$

  • $\begingroup$ Took mu a while to get it, but this worked, thanks! $\endgroup$ – Achifaifa Feb 12 '14 at 14:40

An alternative, no need for partial fractions.

Starting the same way as lab bhattacharjee, note that $x^6 + 1 \equiv (x^2 + 1)(x^4 - x^2 + 1):$

$$\begin{align*} \int \frac{x^4 + 1}{x^6 + 1} dx = \int\frac{x^4 - x^2 + 1 + x^2}{(x^2 +1)(x^4 - x^2 + 1)} dx & = \int\frac{x^4 - x^2 + 1}{(x^4 - x^2 + 1)(x^2 + 1)} dx + \int\frac{x^2}{x^6 + 1} dx \\ & = \int\frac{dx}{x^2 + 1} + \int\frac{x^2}{x^6 + 1} dx \\ & = \arctan x + \tfrac{1}{3} \arctan x^3 + \mathcal{C}\end{align*}$$


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