partial fractions when the fraction cannot be decomposed I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can I then go about writing this fraction as the sum $\frac {1}{2}\left [\frac {1}{x^2+1}-\frac{x^2-1}{(x^2+1)^2}\right]$? Is it just manipulation and trial and error?

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