Q:

What are the criteria for partial fractions?

Consider the example of 

$$1\over (u^2 - 1)(u+1)$$.

$${Au + B\over u^2 - 1} + {C \over u + 1} = {Au^2 + Au + Bu + B + Cu^2 - C\over (u^2 - 1)(u+1)} = {1\over (u^2 - 1)(u+1)}$$
$$\begin {cases}A + C =0 \\ A + B = 0\\ B - C = 1\end{cases}$$
Adding $3$ and $1$,
$$\begin {cases}A + B =1 \\ A + B = 0\end{cases}$$

The degree of denominator is $3$ and that of numerator is $0$, so I should not have got inconsistent group of equations. 
What are the criteria for partial fractions ? what am I missing in this example ?
Also Wolfram Alpha gives proper partial fraction of the following.

A:

Let's change variables to make things clearer (won't affect the theory, of course). If we put $1+u=x$, the original fraction becomes
$$ \frac{1}{((x-1)^2-1)x}  = \frac{1}{(x^2-2x)x} = \frac{1}{x^2(x-2)}. $$
Now, your question boils down to what happens when we add two fractions
$$ \frac{A}{x}+\frac{Bx+C}{x(x-2)}. $$
But we can put them over the common denominator $x(x-2)$, so
$$ \frac{A}{x}+\frac{Bx+C}{x(x-2)} = \frac{Ax-2A+Bx+C}{x(x-2)} = \frac{(A+B)x-2A+C}{x(x-2)}!  $$
So this is the problem: no matter what polynomial we put on top of these fractions, they will never produce a denominator that is $x^2(x-2)$!
Okay, what about
$$ \frac{A}{x^2} + \frac{Bx+C}{x(x-2)}? $$
This looks better, since we have to put it over a common denominator $x^2(x-2)$. Then we get
$$ \frac{Ax-2A+Bx^2+Cx}{x^2(x-2)}. $$
Therefore we're forced to take $B=0$. We also have to get rid of the $x$ term, so $A=-C$. Then you can choose $A$ in order to get $-2A=1$, so
$$ \frac{-1/2}{x^2} + \frac{1/2}{x(x-2)} $$
is a valid partial fraction decomposition for $1/x^2(x-2)$ (and equally,
$$ \frac{-1/2}{(1+u)^2} + \frac{1/2}{u^2-1} = \frac{1}{(u+1)(u^2-1)}). $$

