partial fractions when the fraction cannot be decomposed

It can be decomposed if you use the complex roots $$(x^2+1)^2=(x-i)(x+i)(x-i)(x+i)$$ $$\frac 1{(x^2+1)^2}=\frac{i}{4 (x+i)}-\frac{1}{4 (x+i)^2}-\frac{i}{4 (x-i)}-\frac{1}{4 (x-i)^2}$$

$\frac{1}{(x^2+1)^2}$ is already in "partial fractions form", so...

"Is it just manipulation and trial and error?"

Yes.

It's not super trivial, but the result is true.

I did:

\begin{align} \frac{1}{(x^2+1)^2} &= \frac{1 + x^2 - x^2}{(x^2+1)^2} \\\\ &= \frac{1}{(x^2+1)}\ - \frac{x^2}{(x^2+1)^2} \\\\ &= \frac12\left[\frac{2}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right] \\\\ &= \frac12\left[\frac{1}{(x^2+1)} + \frac{1}{(x^2+1)}\ - \frac{2x^2}{(x^2+1)^2}\right]\\\\ &= \frac12\left[\frac{1}{(x^2+1)}\ + \frac{x^2+1-2x^2}{(x^2+1)^2}\right] \\\\ &= \frac12\left[\frac{1}{(x^2+1)}\ - \frac{x^2-1}{(x^2+1)^2}\right]. \\\\ \end{align}

In the real-valued partial fraction method, your term $$ \frac{1}{(x^2+1)^2} $$ is the best you can do. Evaluation of integrals $$ \int\frac{dx}{(ax^2+bx+c)^n} $$ is done as follows: For $n=0$, complete the square and it is an arctangent. For $n>1$ there is a reduction formula from integration by parts: $$ \int\frac{dx}{(ax^2+bx+c)^n} = \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}} +\frac{(2n-3)2a}{(n-1)(4ac-b^2)} \int\frac{dx}{(ax^2+bx+c)^{n-1}} $$