TITLE: partial fractions when the fraction cannot be decomposed
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Just to clarify:
In cases of irreducible denomiantors over the real numbers such as your case, we treat it like a regular fraction with a repeated factor in the denominator but instead of letting the numerators of each fraction produced by the partial fraction decomposition be constant, instead we let the numerator be a polynomial of degree one lower than the factor in question.
That was rather a mouthful! To clarify, in your case:
$$\frac{1}{(x^2+1)^2}\equiv\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$$
Finding those coefficients, however, yields $D=1$ and this is the only non zero coefficient. This is why your fraction is already in  partial fraction form.