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?

Solutions Collecting From Web of "Which rational functions are derivatives of rational functions?"

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

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.


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.

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.