I was reviewing a classmate (call him Bob)'s work on an integration of a rational expression (although integration is involved, it's beyond the scope of this question).
The problem was:
$$\int\frac{x^2 + 1}{x^2 + 3x + 2} dx$$
Bob attempted to use decomposition to split this into partial fractions. I had been taught that partial fraction decomposition can only be used on a $N(x)/D(x)$ if and only if the $deg(N(x)) < deg(D(x))$, so I knew that Bob's method was incorrect from the start; however, he still got the correct answer.
His work was as follows:
$$\frac{x^2 + 1}{x^2 + 3x + 2} = \frac{A}{x+2} + \frac{B}{x+1} $$
$$ \begin{align} x^2 + 1 &= A(x + 1) + B(x + 2) \\ &= (A + B)x + (A + 2B) \end{align} $$
Comparing the coefficients, he wrote:
$$ \begin{align} 1 &= A + 2B \\ x^2 &= (A + B)x \Rightarrow x = A + B \quad\color{red}{(wrong!)} \end{align} $$
Solving the system, he then wrote $$ \begin{align} 1 &= A + 2B \Rightarrow A = 1-2B\\ x &= A + B \Rightarrow x = (1-2B)+B = 1-B \Rightarrow B=1-x \\ \text{Thus, } A &= 1-2(1-x) = 2x - 1 \end{align} $$
which meant his "decomposition" produced the partial fractions of
$$\frac{x^2 + 1}{x^2 + 3x + 2} = \frac{2x-1}{x+2} + \frac{1-x}{x+1}$$
Although the method was not typical, it ended with fractions that are equivalent to the original, and thus integrates to the correct answer.
Is this just some lucky coincidence, or is there an exception to the rule of when to use partial fraction decomposition?