How can we solve integration for fractions including $x^5$ and $x^7+1$?

Can anyone please solve the integration below? I've been trying it for more than an hour. But no luck.

$$\int \frac{x^5}{x^7+1} dx$$

• "I've been trying for minutes", mathematicians are rolling in their graves right now. ;) – Zach466920 Mar 18 '15 at 20:20
• There's always partial fractions, though I'm sure it's not going to be pretty... – Mike Mar 18 '15 at 20:31
• Basically you can do this with partial fractions. It is easier to find them if you are familiar with complex numbers - roots of unity in particular. The zeros of the denominator are $x=-1$ and then primitive fourteenth roots of unity. Calculating the residues is easy, as all those zeros are simple. Then you pair up conjugate zeros to get the real partial fractions and go from there (or find a complex primitive and start manipulating that). This is no problem for Mathematica/WA. – Jyrki Lahtonen Mar 18 '15 at 20:33
• It just looks like a lot of hard work. The final result can be found here. – Lucian Mar 18 '15 at 20:33
• On a related note, $~\displaystyle\int_0^\infty\frac{x^5}{x^7+1}~dx~=~\frac\pi7~\csc\frac\pi7~$ – Lucian Mar 18 '15 at 20:50

Using power series we can represent, $$\frac{1}{1+x^7}=\frac{1}{1-(-x^7)}= \sum_{n=0}^{\infty}(-x^7)^n$$

$$= \sum_{n=0}^{\infty}(-1)^nx^{7n}=1-x^7+x^{14}-…$$

So $$\frac{x^5}{1+x^7}=x^5-x^{12}+x^{19}-…$$

And now it is a problem if $$\int x^{5}-x^{12}+x^{19}-…dx$$

Other than that you may have wrote the question down wrong, because a closed solution to this problem is huge and will take a very long time to actually get..

• I think this really is the easiest way to do the problem by hand. Partial fractions,according to Lucian's Mathematica solution above in the comments, gets an answer,but it's extremely laborious. – Mathemagician1234 Mar 18 '15 at 20:37
• Your expansion of the fraction is wrong. The exponents increase, not decrease. – marty cohen Mar 18 '15 at 20:49
• @martycohen Thanks, does it look better now? – Quality Mar 18 '15 at 21:00
• Yep. Now, integrate term by term. An alternative is to look up multisection of series, and apply that before integrating. You will see why the trig functions appear. – marty cohen Mar 18 '15 at 21:15

Don't know why you edited your post, but there is a real answer without the "contour detour" as I like to put it. Someone else can prove the result, I'm not interested.

$$\int {{x^5} \over (x^7+1)} dx = -1/7 \sin((3 \pi)/14) \log(x^2+2 x \sin((3 \pi)/14)+1)+{1/7} \sin(\pi/14) \log(x^2-2 x \sin(\pi/14)+1)+1/7 \cos(\pi/7) \log(x^2-2 x \cos(pi/7)+1)-1/7 \log(x+1)+2/7 \sin(pi/7) \tan^{-1}(\csc(\pi/7) (x-\cos(\pi/7)))+2/7 \cos((3 \pi)/14) \tan^{-1}(\sec((3 \pi)/14) (x+\sin((3 \pi)/14)))+2/7 \cos(\pi/14) \tan^{-1}(\sec(\pi/14) (x-\sin(\pi/14)))+C$$ (likelihood of a typo 20%)