# Which rational functions are derivatives of rational functions?

I thought it was interesting that $\frac{u^2+1}{(u^2-2u-1)^2}$ has the very simple integral $-\frac{u}{u^2-2u-1}$ but both of $\frac{u^2}{(u^2-2u-1)^2}$ and $\frac{1}{(u^2-2u-1)^2}$ are very complicated (the transcendental parts cancel each other though).

So my question is how do I check by looking at a rational function whether or not it's a derivative of a rational function?

For example $\frac{1}{(x^2+1)^2}$ isn't but $\frac{x}{(x^2+1)^2}$ is. How can we tell in general?

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Could this be related to checking if the recurrence relations associated to the generating function satisfy $n|a_n$? – user58512 Feb 7 '13 at 21:40
Consider the partial fractions decomposition of a rational function. What kind of terms have rational primitives? And which ones don't? – 1015 Feb 7 '13 at 21:42
See here, over the reals: en.wikipedia.org/wiki/Partial_fractions – 1015 Feb 7 '13 at 21:55
@julien, thank you but does applying partial fractions reduce the problem to something simpler? I can't see it – user58512 Feb 7 '13 at 21:57
Yes, there are only finitely many cases to consider then. For instance, as soon as the decomposition involves $1/(x-\alpha)$, your primitive has some $\ln|1-\alpha|$ and it's not gonna be rational. Likewise, if there is some $1/(x^2+a^2)$ ($a\neq 0$), there will be some $\arctan (x/a)$ and the primitive will not be rational. – 1015 Feb 7 '13 at 21:59

Examine the poles of your function (in the complex plane). If all residues are zero, you are in good shape.

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thank you very much. is this an if and only if condition? – user58512 Feb 7 '13 at 22:32
Of course. You will learn about this when you study complex analysis. – GEdgar Feb 8 '13 at 1:29

For your last example, it is easy to see that a) you have a simple factor of $x$ in the numerator, and b) your denominator is a simple power of $1+x^2$. The integral is easily transformed into the form $\int du/(1+u)^2$, which is a rational function. This works for any power of $1+x^2$ greater than 1.

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To make a little more clear the reason why GEdgar's solution is works, observe that since all polynomials factor over $\mathbb{C}$, you can take the rational function $r$ and expand it completely using the method of partial fractions.

That is, if $r$ has poles at $z_1,z_2,\ldots,z_k$ with multiplicities $m_1,m_2,\ldots,m_k$ respectively, then for each $i\in\{1,\ldots,k\}$, and each $j\in\{1,\ldots,m_i\}$, there is are constants $a_{ij}$ such that the following holds.

$$r(z)=\dfrac{a_{11}}{z-z_1}+\dfrac{a_{12}}{(z-z_1)^2}+\cdots+\dfrac{a_{1m_1}}{(z-z_1)^{m_1}}+\cdots+\dfrac{a_{k1}}{z-z_k}+\dfrac{a_{k2}}{(z-z_k)^2}+\cdots+\dfrac{a_{km_k}}{(z-z_k)^{m_k}}.$$

Every term here is integrable, except the terms $\dfrac{a_{i1}}{z-z_1}$ where $a_{i1}\neq0$. This $a_{i1}$ is exactly the residue of the function at $z_i$. Therefore we get the criterion described by GEdgar.

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