<s>[INST] Is this a valid partial fraction decomposition? 
Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions.
$$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$
$$4x+1 = A(x-2)+B(x+1)$$
$$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$
$$x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1$$
Thus,
$$\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}$$

The substitution of $x$ ($x = 2, -1$) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying $(x+1)(x-2)$, which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at $x=-1,2$.
How can we substitute these value for $x$? So is this method valid and rigorous? How to modify it so that it is rigorous?
 [/INST] $A(x-2)+B(x+1) = 4x+1$ gives us a system of two equations:
$$\begin{split}
A+B &= 4 \\
-2A + B &= 1
\end{split}$$
But really the first equation as really $(A+B)x = 4x$, so the substitution $x=2$ simply gives us the system:
$$\begin{split}
2A+2B &= 8 \\
-2A + B &= 1 \\
\end{split}$$
Which conveniently sums to $3B = 9$, $B=3$. It's just a choice of substitution that makes it easy to solve the system. Any such substitution is valid, just that $x=2$ and $x=-1$ happen to be particularly good choices. 
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