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?
 A: It can be decomposed if you use the complex roots
$$(x^2+1)^2=(x-i)(x+i)(x-i)(x+i)$$
$$\frac 1{(x^2+1)^2}=\frac{i}{4 (x+i)}-\frac{1}{4 (x+i)^2}-\frac{i}{4 (x-i)}-\frac{1}{4 (x-i)^2}$$
A: $\frac{1}{(x^2+1)^2}$ 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}
A: 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 $n=1$, complete the square and it is an arctangent.  For $n>1$ there is a reduction formula from integration by parts:
\begin{align}
\int\frac{dx}{(ax^2+bx+c)^n} &= \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}
\\ &\qquad+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}
\int\frac{dx}{(ax^2+bx+c)^{n-1}}
\end{align}
A: 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.
