Determining which values to use in place of x in functions When solving partial fractions for integrations, solving x for two terms usually isn't all that difficult, but I've been running into problems with three term integration.
For example, given 
$$\int\frac{x^2+3x-4}{x^3-4x^2+4x}$$
The denominator factored out to $x(x-2)^2$, which resulted in the following formulas
$$
\begin{align*}
x^2+3x-4=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}\\
x^2+3x-4= A(x-2)(x-2)^2+Bx(x-2)^2+Cx(x-2)\\
x^2+3x-4= A(x-2)^2+Bx(x-2)+Cx\\\\
\text{when x=0, }A=-1 
\text{ and x=2, }C=3
\end{align*}
$$
This is where I get stuck, since nothing immediately pops out at me for values that would solve A and C for zero and leave some value for B.  How do I find the x-value for a constant that is not immediately apparent?

The point of these equations involving $A$, $B$, $C$, and $x$ is that there are unique values of $A$, $B$, and $C$ that make the decomposition work, but they are supposed to be valid for every value of $x$. In particular, they should be true when $x = 0$ and $x = 2$. These values happen to be convenient because they cause all but one of the unknowns to vanish, thus allowing you to easily solve for the remaining unknown. Once you know $A$ and $C$, however, your equation looks like 
$$
\begin{align*}
x^2 + 3x - 4 = -(x-2)^2 + Bx(x-2) + 3x,
\end{align*}
$$
since $A = -1$ and $C = 3$ are the unique solutions. I re-emphasize that they do not depend on the choice of $x$; we just chose convenient values for $x$ to help us discover $A$ and $C$.
To finish, we can just use any value for $x$, leaving $B$ as the only unknown. I'll choose $x = 1$, simply because I'm not very imaginative:
$$
\begin{align*}
1^2 + 3 \cdot 1 - 4 &= -(1 - 2)^2 + B \cdot 1 (1 - 2) + 3 \cdot 1\\
0 &= -B + 2\\
B &= 2.
\end{align*}
$$
