$\int \frac{dx}{(x^4 + 1)^2}$

Well, the problem is tagged "partial fractions," that seems to me to be the way to go. Of course, first you need to factor $x^4+1$ into irreducible quadratics; that can be done by $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2$$

It will be messy - do you have any reason to think there is a simpler method?

Edit: for the definite integral, if you've done complex variables, you'll know about contour integration, which ought to handle this example without difficulty. If you haven't done complex variables, you have something truly wonderful to look forward to.

The definite integral can be done using residues. Note that the zeros of $(z^4+1)^2$ in the upper half plane are at $\frac{\pm 1 + i}{\sqrt{2}}$, both with multiplicity 2. The residues are $\frac{3 \sqrt{2}}{32}(1-i)$ at $\frac{-1+i}{\sqrt{2}}$ and $\frac{3 \sqrt{2}}{32}(-1-i)$ at $\frac{1+i}{\sqrt{2}}$ (note that the residue of $\frac{1}{((x-r) f(x))^2}$ at $x=r$, where $f$ is analytic at $r$ with $f(r) \ne 0$, is $\frac{-2 f'(r)}{f(r)^3}$). So the integral is $2 \pi i \frac{3 \sqrt{2}}{32} (1-i -1 - i) = \frac{3 \pi \sqrt{2}}{8}$.