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