= Solution
Use <partial fraction decomposition>:
$$
\frac1{x^2(x^2-4)}=-\frac1{4x^2}+\frac1{4(x^2-4)}
=-\frac1{4x^2}+\frac1{16}\left(\frac1{x-2}-\frac1{x+2}\right).
$$
Integrating each term gives the <antiderivative>
$$
\boxed{\int\frac{dx}{x^2(x^2-4)}=\frac1{4x}+\frac1{16}\log\left|\frac{x-2}{x+2}\right|+C.}
$$
This holds on any interval avoiding $-2,0,2$; the constant may differ between these intervals. <Differentiation> of the logarithm recovers $1/[4(x^2-4)]$, providing a direct check.
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