# Thread: Integration by Partial Fractions

1. ## Integration by Partial Fractions

This problem is taken from Stewart, A58, no.46.

This is under the section: Integration of Rational Fuctions by Partial Fractions.

If f is a quadratic function such that f(0) = 1, and
$\displaystyle \int$f(x).dx/[x^2.(x+1)^3]
is a rational function, find the value of f'(0).

Sorry, but I am having trouble putting it into latex.Any help is much appreciated.

2. $\displaystyle f(x)$ is a quadratic function. We can say that $\displaystyle f(x) = a x^2 + bx + c$
If $\displaystyle f(0) = 1$, we find that $\displaystyle c = 1$
So we can rewrite $\displaystyle f(x)$ as $\displaystyle ax^2 + bx + 1$

We want to find $\displaystyle f'(0)$.
$\displaystyle f(x) = ax^2 + bx + 1$
$\displaystyle f'(x) = 2ax + b$
$\displaystyle f'(0) = b$
That means we are looking for $\displaystyle b$.

$\displaystyle \int \frac{ax^2 + bx + 1}{x^2(x+1)^3}dx$

You can expand it and then integrate it using partial fraction expansions. It's a long work and a real pain to write it here. I assume that you find it correctly. It is,

$\displaystyle \frac{b-a-1}{2(x+1)^2}-\frac{1}{x}+\frac{b-2}{x+1}+(b-3)\ln x + (b-3)\ln (x+1)$

$\displaystyle \frac{x^2(2b - 6) + x(3b - a - 9 ) - 2}{2x(x+a)^2}+(b-3)\ln x + (b-3)\ln (x+1)$

$\displaystyle b = 3$ removes the logarithm functions and leaves us a rational function

3. Aha! That's genius! I got up to that last line of latex, then didn't know what to do. I agree it is a lot of working out. Too much, in fact. I wonder if there is a simpler method of working it out.
Thank-you, I really never saw that.