Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$ Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral:
$$\int\frac{dx}{(x^4-16)^2}$$
I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when regrouping the coefficients for the powers of $x$ and then solving an $8\times8$ system of linear equations. It will eventually work, but I suspect there is something more ingenuine behind this problem.
I also tried all sorts of trigonometric substitutions and formulations, but that added square power really is a bummer to it all.
I'm generally open to any exchange on the topic and would be glad to hear some advice in such situations. Many thanks in advance!

Just integrate by parts
\begin{align}
\int\frac{dx}{(x^4-16)^2}
& =\int \frac1{64x^3} d\left( \frac{-x^4}{x^4-16}\right)
=-\frac1{64} \frac x{x^4-16} -\frac3{64} \int \frac{dx}{x^4-16}\\
&= -\frac1{64} \frac x{x^4-16} -\frac3{64}\cdot\frac18\int \left( \frac1{x^2-4}-\frac1{x^2+4} \right)dx
\end{align}
