# 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.

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You can't use partial fractions without knowing the expression for $f(x)$. –  student Feb 10 '12 at 14:54
$f(x)=ax^2+bx+c$ –  pedja Feb 10 '12 at 14:55
@pedja: oh, I missed that. Btw, I edited the question before and my edit was erased by you. Was there any problem? –  student Feb 10 '12 at 14:57
@Leandro,I didn't see your edit... –  pedja Feb 10 '12 at 14:58
Using that $f(0) = 1$, we can rewrite $f(x) = ax^2 + bx + 1$. The derivative of this is $f ' (x) = 2ax + b$. Hence $f'(0) = b$. –  The Chaz 2.0 Feb 10 '12 at 14:58

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, $f'(0)=3$.

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Thank you for taking the time to write out the solution, makes perfect sense. –  Leslie Feb 10 '12 at 16:02
@Leslie: In answer to your offline question, do you see the up and down arrows to the left at the top of David's question? Below that should be a grayed out check mark. As the questioner, you should be able to click that check mark and turn it green, accepting David's answer. –  Jeff Feb 10 '12 at 23:49
@ Jeff, thanks. I got it! –  Leslie Feb 11 '12 at 0:54