How to integrate $\int \frac{x^{13}\ dx}{x^5 + 1}$

Question

We get this problem from our teacher today. I only wish that it was $x^{14}$ in the numerator, so we can use substitution method:

$$\int \dfrac{x^{13}\ dx}{x^5 + 1}$$

I can't find way to integrate this. Please help me out to solve this seeming simple integral.

Answer 1

$$ \int\frac{x^{13}}{x^5+1}\,dx =\int\left(x^8-x^3+\frac{x^{3}}{x^5+1}\right)\,dx =\frac{x^9}{9}-\frac{x^4}{4}+\int\frac{x^{3}}{x^5+1}\,dx $$ Note that $$ x^5+1=(x+1)(x^2+\phi x+1)(x^2-(\phi-1) x+1), $$ where $\phi=\frac{\sqrt{5}+1}{2}$ is the "Golden Ratio". It can be shown that $$ \frac{x^3}{x^5+1}=\frac{1}{5}\left(\frac{\phi x+1-\phi}{x^2-(\phi-1)x+1}+ \frac{(1-\phi)x+\phi}{x^2+\phi x+1}-\frac{1}{x+1}\right) $$ You can take it from here...

Answer 2

E.g. with $\frac{x^{13}}{x^5+1}=\sum\limits_{k=0}^\infty (-1)^k x^{5k+13}$ you can integrate easily for $|x|<1$.

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You can simplify with $x^{13}=(x^8-x^3)(x^5+1)+x^3$.

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The useful link of user 170039 gives you the result for $\int\frac{x^{3}}{x^5+1}$ and shows how complicated it becomes.

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If you like to use the proposal of boaz, then you need $\int\frac{ex+f}{ax^2+bx+c}=$
$$=\frac{e}{2a}\ln|ax^2+bx+c|+\frac{2af-be}{a\sqrt{4ac-b^2}}\cdot\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}+C$$ for $4ac-b^2>0$ and $$=\frac{e}{2a}\ln|ax^2+bx+c|+\frac{2af-be}{2a\sqrt{b^2-4ac}}\cdot\ln\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}+C$$ for $4ac-b^2<0$.