Why don't we have to use polynomial long division on quadratic irreducible partial fractions in this case? For $(x^2+3x+1)
/(x^2+1)^2 $
Why don't we have to use polynomial long division in this case since the degree of the denominator has the same magnitude of degree as the numerator before doing partial fraction decomposition?
 A: You can use long division in this particular case, where the denominator is the power of a quadratic. But it is not much help in other cases.
\begin{eqnarray}
\frac{x^2+3x+1}{(x^2+1)^2}&=&\frac{x^2+3x+1}{x^2+1}\cdot\frac{1}{x^2+1}\\
&=&\left(1+\frac{3x}{x^2+1}\right)\cdot\frac{1}{x^2+1}\\
&=&\frac{1}{x^2+1}+\frac{3x}{(x^2+1)^2}
\end{eqnarray}
ADDENDUM: Here is another demonstration of the technique:
Decompose $\dfrac{x^4+1}{(x^2+x+1)^3}$
\begin{eqnarray}
\frac{x^4+1}{(x^2+x+1)^3}&=&\left(\frac{x^4+1}{x^2+x+1}\right)\cdot\frac{1}{(x^2+x+1)^2}\\
&=&\left(x^2-x+\frac{x+1}{x+x+1}\right)\cdot\frac{1}{(x^2+x+1)^2}\\
&=&\left(\frac{x^2-x}{x^2+x+1}+\frac{x+1}{(x^2+x+1)^2}\right)\cdot\frac{1}{x^2+x+1}\\
&=&\left(1-\frac{2x+1}{x^2+x+1}+\frac{x+1}{(x^2+x+1)^2}\right)\cdot\frac{1}{x^2+x+1}\\
&=&\frac{1}{x^2+x+1}-\frac{2x+1}{(x^2+x+1)^2}+\frac{x+1}{(x^2+x+1)^3}
\end{eqnarray}
A: We don't have to, we can go backwards, make ansatz, multiply by least common multiple for denominator and solve linear equation system that arises.
$$\frac{a}{x^2+1} + \frac{bx+c}{(x^2+1)^2}$$
Now multiply both numerator and denominator of first term by $x^2+1$
$$\frac{a(x^2+1) + bx+c}{(x^2+1)^2}$$
Now set up equations for what $a$, $b$ and $c$ are. Can you finish from here?
