# Thread: A Tricky Problem - Need Help Seeing The Trick Please

1. ## A Tricky Problem - Need Help Seeing The Trick Please

If f is a quadratic function such that f(0) =1 and $\int\frac{f(x)dx}{x^2(x+1)^3}$ is a rational function. Find the value of f'(0).

I'm running out of ideas, anyone help pls? I think the derivative of the denominator = f(x), is that correct? Please put solution to the entire problem so I can check if I'm doing things right. Thanks.

2. Originally Posted by markxchan
If f is a quadratic function such that f(0) =1 and $\int\frac{f(x)dx}{x^2(x+1)^3}$ is a rational function. Find the value of f'(0).

I'm running out of ideas, anyone help pls? I think the derivative of the denominator = f(x), is that correct? Please put solution to the entire problem so I can check if I'm doing things right. Thanks.
$f(x)$ is quadratic such that $f(0) = 1$ ...

$f(x) = ax^2 + bx + 1$

$f'(x) = 2ax + b
$

$f'(0) = b$

$\int \frac{ax^2 + bx + 1}{x^2(x+1)^3} \, dx$ is a rational function ...

using the method of partial fractions with repeated linear factors ...

$\frac{ax^2 + bx + 1}{x^2(x+1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3}$

since the integral is rational, A and C must equal 0 since their antiderivatives would yield a log function.

$\frac{ax^2 + bx + 1}{x^2(x+1)^3} = \frac{B}{x^2} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3}$

$ax^2 + bx + 1 = B(x+1)^3 + Dx^2(x+1) + Ex^2$

$ax^2 + bx + 1 = (B+D)x^3 + (3B+D+E)x^2 + (3B)x + B
$

equating coefficients ...

$B+D = 0$

$3B+D+E = a$

$3B = b$

$B = 1$

hence ... $b = 3$ and $f'(0) = b = 3$