Calculus 2: Partial fractions problem. Finding the value of a constant I encountered the following problem. 
Let $f(x)$ be a quadratic function such that $f(0) = -6$ and
$$\int \frac{f(x)}{x^2(x-3)^8} dx $$
is a rational function.
Determine the value of $f'(0)$
Here's what I tried. I decomposed the fraction integrand below
$$\frac{f(x)}{x^2(x-3)^8} = \frac{A_1}{x} + \frac{A_2}{x^2} +\sum_{i=1}^8 \frac{B_i}{(x-3)^i}$$
By finding a common denominator, I determined 
$$f(x) = A_1x(x-3)^8 + A_2(x-3)^8 +x^2 \sum_{j=1}^8 [B_j(x-3)^{8-j} ]$$
$$f'(x) = A_1(x-3)^8 + 8A_2(x-3)^7 + D(x) $$
where $D(x)$ is a function such that $D(0) = 0$. (These are all the remaining terms that go away when we plug $0$ into $f'(0)$).
I used the information that $f(0) = -6$ to get the equation
$$f(0) = A_2(-3)^8 = -6 \rightarrow A_2 = \frac{-2}{3^7}$$. 
This leaves us with 
$$f'(0) = A_1(-3)^8 +8\cdot \frac{-2}{3^7} \cdot (-3)^7 $$
$$ f'(0)= 3^8 \cdot A_1 + 16$$
I was told $f'(0) = 16$, however I cannot convince myself the value of $A_1$. I would think that the fact that $f(x)$ is a quadratic function should come into play here. Please let me know what you think.
 A: Note that since $f$ is quadratic with $f(0)=0$, we can write $f(x)$ as $$f(x)=-6+f'(0)x+\frac12 f''(0)x^2$$
Then, the integrand becomes

$$\frac{f(x)}{x^2(x-3)^8}=\frac{-6}{x^2(x-3)^8}+\frac{f'(0)}{x(x-3)^8}+\frac{\frac12 f''(0)}{(x-3)^8} \tag 1$$

The last term on the right-hand side of $(1)$ integrates to the rational function, $\int \frac{\frac12 f''(0)}{(x-3)^8}\,dx =-\frac12f''(0)\frac{1}{7(x-3)^7}+C$ and is not implicated , therefore, in the ensuing analysis.
Using partial fraction expansion, we can write the first term on the right-hand side of $(1)$ as 
$$\frac{-6}{x^2(x-3)^8}=\frac{-16}{6561\,x}+\frac{16}{6561\,(x-3)}+\left(\text{other terms that integrate to rational functions}\right) \tag 2$$
and the second term on the right-hand side of $(1)$ as
$$\frac{f'(0)}{x(x-3)^8}=\frac{f'(0)}{6561\,x}-\frac{f'(0)}{6561\,(x-3)}+\left(\text{other terms that integrate to rational functions}\right) \tag 3$$
Therefore, to annihilate the terms that integrate to $\log(x)$ and $\log(x-3)$ in $(2)$ and $(3)$, we must have $$f'(0)=16$$
A: Consider that your partial fraction decomposition  is equal to $\frac {A_1}{x}+\frac {B_1}{x-3}+h(x),$ where $h(x)$ is the sum of the other terms. The anti-derivative of each of those other terms is a rational function. But $\int (\frac {A_1}{ x}+\frac {B_1}{x-3})\;dx=A_1\ln |x|+B_1\ln |x-3|$ is not a rational function unless $A_1=B_1=0.$ So $A_1=0$ and you are finished.
