trouble solving a partial fraction decomposition problem with a numerator of 1 and two irreducible quadratic factors in denominator I am stuck on a partial fraction decomposition problem. $$\frac{1}{u^{2} +1}\frac{1}{a^{2}u^{2}+1} = \frac{a^{2}}{(a^{2}-1)(a^{2}u^{2}+1)}-\frac{1}{(a^{2}-1)(u^{2}+1)}$$ I can't seem to come up with what is on the right side of the equal sign. I know there are 2 irreducible quadratic factors in the denominator, but when I try $Ax+B$, $Cx+D$ and do the cross multiplication I end up getting stuck on setting up my set of equations. Any help would be appreciated.
 A: Hint:  if you consider $u^2$ as the variable then the problem can be solved as a partial fraction decomposition in $x=u^2$, just determine $A,B$ such that: $$\frac{1}{(x+1)(a^2x+1)}=\frac{A}{x+1}+\frac{B}{a^2x+1}$$
A: In general, if you have a rational function $\frac{P(x)}{Q(x)}$ with $\deg P < \deg Q$ and the roots $\lambda_1, \ldots, \lambda_q$ of $Q(x)$ are distinct, you have following partial fraction decomposition
$$\frac{P(x)}{Q(x)} = \sum_{i=1}^q \frac{P(\lambda_i)}{Q'(\lambda_i)(x-\lambda_i)}$$
In the special case $P(x) = 1$ and $Q(x)$ is a product of factors, this reduces to
$$\prod_{i=1}^q \frac{1}{(x-\lambda_i)} = \sum_{i=1}^q \frac{1}{x-\lambda_i}\prod_{j=1,\ne i}^q \frac{1}{\lambda_i - \lambda_j}$$
For the expression at hand, you can view it as a rational function in $u^2$. 
Using above constructions, you can read off following decomposition of your expression.
$$\begin{align}
\frac{1}{(u^2+1)(a^2u^2+1)} 
&= \frac{1}{(u^2+1)(a^2(-1)+1)} + \frac{1}{(-(1/a^2)+1)(a^2u^2+1)}\\
&= -\frac{1}{(u^2+1)(a^2-1)} + \frac{a^2}{(a^2-1)(a^2u^2+1)}
\end{align}$$
When the poles are distinct, you should use this approach instead of matching coefficients to figure out the decomposition.
A: First of all, let's assume that $a\neq1$, because this value would create a different type of denominator that would require setting up different partial fraction decomposition. (And we might as well assume that $a$ is positive, since it's squared anyway.)
You correctly described the standard approach here. So if we continue with it, you should set up the equation
$$\frac{1}{u^2+1}\cdot\frac{1}{a^2u^2+1}=\frac{Au+B}{u^2+1}+\frac{Cu+D}{a^2u^2+1}$$
with undetermined coefficients. Multiplying both sides by the original denominator $(u^2+1)(a^2u^2+1)$, we get
$$1=(Au+B)(a^2u^2+1)+(Cu+D)(u^2+1).$$
Multiplying out an collecting the like terms on the right-hand side, we get
$$1=(a^2A+C)u^3+(a^2B+D)u^2+(A+C)u+(B+D).$$
These two polynomials must be identically equal, so their respective coefficients must be the same. Note that there are two coefficients involving only $A$ and $C$ and two coefficients involving only $B$ and $D$. So we can set up two separate systems of two equations with two unknowns:
$$a^2A+C=0,A+C=0 \quad \text{and} \quad a^2B+D=0,B+D=1.$$
Solving these will give you the desired answer.
Note 1. This is not the shortest or the most efficient solution for this particular question. But this is a good exercise in following the standard approach for partial fraction decomposition. If you observe the special form of the fractions, then you can arrive at the same answer much faster, as shown above by @divx, with the solution he suggested in the comment being the most elegant, imho.
Note 2. What about the special case? If $a=1$, then the denominator has a repeated irreducible quadratic factor (rather than two different ones). And in fact, the fraction
$$\frac{1}{(u^2+1)^2}$$
already is a partial fraction, so it can't be decomposed any further.
