$\int \frac{dx}{(x^4 + 1)^2}$ What would be a relatively simple method for computing the indefinite integral below? 

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

Furthermore, how would one evaluate the following, possibly by detouring the computation of the indefinite integral?

$\displaystyle\int\nolimits_{-\infty}^\infty \frac{dx}{(x^4+1)^2}$

Note: The resolved value of the definite integral (according to wolfram alpha) is $\displaystyle \frac{3\pi}{4\sqrt{2}}$

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 $2 \pi i \frac{3 \sqrt{2}}{32} (1-i -1 - i) = \frac{3 \pi \sqrt{2}}{8}$.  
