Why is numerator re-defining necessary for proper solution of some partial fraction integrations? In my homework, one problem was the following formula.  Using standard partial fraction techniques where you'll see my work, I came up with an almost correct answer in the fact the book solution had a three term integral rather than the two term I used.  The only difference between my work and the correct solution was I didn't rewrite the numerator.  My two-part question is


*

*Why did excluding the numerator re-defining result in the wrong answer?

*Under what circumstances should I use numerator re-defining?


Given the following formula, integrate using partial fractions:
$$\int\frac{2x^3-4x^2-15x+5}{x^2-2x-8}$$
$$
\begin{align*}
2x^3-4x^2-15x+5=\frac{A}{x-4}+\frac{B}{x+2}\\
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
\end{align*}
$$
Let $Ax$=-2
$$
\begin{align*}
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
2(-2)^3-4(-2)^2-15(-2)+5=A(-2+2)+B(-2-4)\\
3=-6B\\
B=-\frac{1}{2}
\end{align*}
$$
Let $Bx= 4$
$$
\begin{align*}
2x^3-4x^2-15x+5=A(x+2)+B(x-4)\\
2(4)^3-4(4)^2-15(4)+5=A(4+2)+B(4-4)\\
9=6A\\
A=\frac{3}{2}
\end{align*}
$$
Using $A \text{ and } B$ values, plug them into the integral as
$$
\begin{align*}
\frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}\\
=\frac{3}{2}\text{ln}\left |x-4 \right |-\frac{1}{2}\text{ln}\left |x+2\right |+C
\end{align*}
$$
This work matches up with the solution process except the original integral was re-written as 
$$\int2x+\frac{x+5}{(x-4)(x+2)}$$
which resulted in the final integral terms
$$\int2x\text{dx}+\frac{3}{2}\int\frac{1}{x-4}\text{dx}-\frac{1}{2}\int\frac{1}{x+2}\text{dx}$$

Normally one uses the division algorithm to decompose into integral and proper fractional parts
$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ =\ 2\:x + \dfrac{x+5}{(x+2)\:(x-4)} $$
Then one performs the partial fraction decomposition only on the second fractional part. But you seem to desire to skip the initial division algorithm step and, instead, jump immediately to the partial fraction decomposition, proceeding essentially as follows.
$$\rm \dfrac{2x^3-4x^2-15x+5}{(x+2)\:(x-4)}\ = \ \dfrac{A}{x-4}\ +\ \dfrac{B}{x+2}\ +\ \: C(x) $$
$$\rm \Rightarrow\ \ \  2\:x^3-4\:x^2-15\:x+5\ = \ A\ (x+2) + B\ (x-4)\ +\ C(x)\:(x+2)\:(x-4) $$
Using the Heaviside cover-up method will still work to find the correct values of $\rm\:A\:$ and $\rm\:B\:$ because the term  $\rm\:C(x)\:(x+2)\:(x-4)\:$ plays no role - it vanishes when evaluating the RHS at $\rm\:x = 4\:$ and $\rm\:x = -2\:$ while solving for $\rm\:A\:$ and $\rm\:B\:.\:$ Similarly  for the LHS which is $\rm\:x+5 + 2\:x\:(x+2)\:(x-4)\:.\:$ Therefore you'll get the same equations for $\rm\:A\:$ and $\rm\:B\:$ whether you use this method or the standard method. Thus, as you do, one can safely set $\rm\:C(x) = 0\:$ from the start. But that yields more work than the standard way, since then you have to evaluate a cubic (vs. $\rm\:x+5\:$) on the LHS. Finally, to obtain the complete answer, you still have to calculate the "integral" part $\rm\:C(x) = 2\:x\:$ and add it to  the partial fraction decomposition. The failure to do so explains why your answer is erroneous. You could get the integral part by subtracting the fractional part from the original expression, but that too would be more work than using the division algorithm from the start.
