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?

To find the constants in the rational fraction $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$
you may use any set of 3 values of $x$, provided that the denominator  $x^3-4x^2+4x\ne 0$.
The "standard" method is to compare the coefficients of $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$ after multiplying this rational fraction by the denominator $x^3-4x^2+4x=x(x-2)^2$ and solve the resulting linear system in $A,B,C$. Since 
$$\begin{eqnarray*}
\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} &=&\frac{A}{x}+\frac{B}{x-2}+%
\frac{C}{\left( x-2\right) ^{2}} \\
&=&\frac{A\left( x-2\right) ^{2}}{x\left( x-2\right) ^{2}}+\frac{Bx\left(
x-2\right) }{x\left( x-2\right) ^{2}}+\frac{Cx}{x\left( x-2\right) ^{2}} \\
&=&\frac{A\left( x-2\right) ^{2}+Bx\left( x-2\right) +Cx}{x\left( x-2\right)
^{2}} \\
&=&\frac{\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A}{x\left(
x-2\right) ^{2}},
\end{eqnarray*}$$
if we equate the coefficients of the plynomials
$$x^{2}+3x-4\equiv\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A,$$
we have the system 
$$\begin{eqnarray*}
A+B &=&1 \\
-4A-2B+C &=&3 \\
4A &=&-4,
\end{eqnarray*}$$
whose solution is
$$\begin{eqnarray*}
B &=&2 \\
C &=&3 \\
A &=&-1.
\end{eqnarray*}$$
Alternatively you could use the method indicated in parts A and B, as an example. 
A.  We can multiply $f(x)$ by $x=x-0$ and $\left( x-2\right) ^{2}$ and let $%
x\rightarrow 0$ and $x\rightarrow 2$. Since $$\begin{eqnarray*}
f(x) &=&\frac{P(x)}{Q(x)}=\frac{x^{2}+3x-4}{x^{3}-4x^{2}+4x} \\
&=&\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} \\
&=&\frac{A}{x}+\frac{B}{x-2}+\frac{C}{\left( x-2\right) ^{2}},\qquad (\ast )
\end{eqnarray*}$$
if we multiply $f(x)$ by $x$ and let $x\rightarrow 0$, we find $A$: $$A=\lim_{x\rightarrow 0}xf(x)=\lim_{x\rightarrow 0}\frac{x^{2}+3x-4}{\left(
x-2\right) ^{2}}=\frac{-4}{4}=-1.$$ And we find $C$, if we multiply $f(x)$ by $\left( x-2\right) ^{2}$ and let $x\rightarrow 2$: $$C=\lim_{x\rightarrow 2}\left( x-2\right) ^{2}f(x)=\lim_{x\rightarrow 2}\frac{%
x^{2}+3x-4}{x}=\frac{2^{2}+6-4}{2}=3.$$
 B. Now observing that $$P(x)=x^{2}+3x-4=\left( x+4\right) \left( x-1\right)$$ we can find $B$ by making $x=1$ and evaluate $f(1)$ in both sides of $(\ast)$, with $A=-1,C=3$: $$f(1)=0=2-B.$$ So $B=2$. Or we could make $x=-4$ in $(\ast)$ $$f(-4)=0=\frac{1}{3}-\frac{1}{6}B.$$ We do obtain $B=2$. 
Thus $$\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}}=-\frac{1}{x}+\frac{2}{x-2}+\frac{3%
}{\left( x-2\right) ^{2}}\qquad (\ast \ast )$$
Remark: If the denominator has complex roots, then an expansion as above is
not possible. For instance
$$\frac{x+2}{x^{3}-1}=\frac{x+2}{(x-1)(x^{2}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{%
x^{2}+x+1}.$$
You should find $A=1,B=C=-1$: 
$$\frac{x+2}{x^{3}-1}=\frac{1}{x-1}-\frac{x+1}{x^{2}+x+1}.$$
