<s>[INST] problem with partial fraction decomposition I want to do partial fraction decomposition on the following rational function:
$$\frac{1}{x^2(1+x^2)^3}$$
So I proceed as follows:
$$\begin{align}
\frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 + x^2)^3} \\
1 = &Ax(1+x^2)^3 + B(1+x^2)^3 + (Cx + D)x^2(1+x^2)^2 \\
& + (Ex + F)x^2(1+x^2) + (Gx + H)(x^2) \\
1 = &(A+C)x^7 + (B+D)x^6 + (3A+2C+E)x^5 + (3B+2D)x^4 \\
& + (3A+C+E+G)x^3 + (3B+D+H)x^2 +(A)x + (B)\\
\end{align}$$
But from this, $B=1$, $B=-D$, and $3B+2D=0$, which is an inconsistent system. What am I doing wrong?
 [/INST] Not answering your question, but I see $x^2$ everywhere, so it is better to find a partial fractions expression for $\dfrac{1}{u(1+u)^3}$.
So we are looking for $A$, $B$, $C$, $D$ such that
$$\frac{1}{u(1+u)^3}=\frac{A}{u}+\frac{B}{1+u}+\frac{C}{(1+u)^2}+\frac{D}{(1+u)^3}.$$
When we bring the right-hand side to a common denominator $u(1+u)^3$, the numerator becomes 
$$A(1+u)^3+Bu(1+u)^2+Cu(1+u)+Du.$$
This is identically equal to $1$. Put $u=0$. We get $A=1$. Put $u=-1$. We get $D=-1$.  The coefficient of $u^3$ is $A+B$ so $B=-1$. The coefficient of $u^2$ is $3A+2B+C$, so $C=-1$. 
Now replace $u$ by $x^2$. 
</s>