Confused by Calc II question regarding derivative of rational integrals So here's the question:
If $f$ is a quadratic function such that $f(0) = 1$ and  $\int \frac{f(x)}{x^2(x+1)^3}\,dx$ is a rational function, find the value of $f’(0)$. 
What I've done so far is try to solve the integral using partial fractions i.e.
$\frac{f(x)}{x^2(x+1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3}$
Multiply out the denominator from the LHS to get:
$f(x) = Ax(x+1)^3 + B(x+1)^3 + Cx^2(x+1)^2 + Dx^2(x+1) + Ex^2$
when $x = 0$ I get that $B=1$.
At this point I'm stuck. I tried solving for the other variables but it gets insanely complicated. Wondering if anyone has a better strategy to solving the problem.
Thank you.
 A: You have
$f(x)=ax^2+bx +c$. That $f(0)=1$ gives you $c=1$. We have $f'(x)=2ax+b$; and so $f'(0)=b$. 
The integrand can be written as
$$
 {ax^2+ f'(0)x+1\over x^2(x+1)^3} = {A\over x\vphantom{ )^2}}+{B\over x^2\vphantom{ )^2}}+ {C\over (x+1)\vphantom{ )^2}}+{D\over (x+1)^2}+{E\over (x+1)^3}.
$$
Here's the important observation:
If the antiderivative of the above is a rational function, then $A=C=0$ (otherwise,  it will contain logarithms).
Thus,
$$
 {ax^2+ f'(0)x+1\over x^2(x+1)^3} = {B\over\vphantom{(^2} x^2}+ {D\over (x+1)^2}+{E\over (x+1)^3};
$$
or,
$$
 {ax^2+ f'(0)x+1 } = {B }(x+1)^3+ {D }x^2(x+1)+{E }x^2.
$$
Setting $x=0$ in the above gives you $B=1 $.
Setting  $x=-1$ in the above gives you $E=a-f'(0)+1$.
Also, comparing the $x^3$ terms, $B=-D$.
So:
$$
\eqalign{
& {    ax^2+ f'(0)x+1 }\ =\ (x+1)^3-  x^2(x+1)+{ (a-f'(0)+1)}x^2\cr 
  \iff&  \color{maroon}{ax^2}+ f'(0)x+1\  =\ (\color{darkgreen}{x^3}+\color{darkblue}{3x^2}+3x+1)  \color{darkgreen}{-x^3}\color{darkblue}{-x^2}+ \color{maroon}{ax^2} +({1-f'(0))}x^2 \cr
  \iff&{ \hphantom{ax^2+} f'(0)x+1 }\ =\  2x^2+3x+1+ \bigl (1 -f'(0)\bigr)x^2 \cr
  \iff&
 {  \hphantom{ax^2+} f'(0)x+1 }\ =\    \bigl(3 -f'(0)\bigr)x^2 +3x+1;
}
$$
whence, $f'(0)=3$. 
