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

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    A Tricky Problem - Need Help Seeing The Trick Please

    If f is a quadratic function such that f(0) =1 and $\displaystyle \int\frac{f(x)dx}{x^2(x+1)^3}$ 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.
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    Quote Originally Posted by markxchan View Post
    If f is a quadratic function such that f(0) =1 and $\displaystyle \int\frac{f(x)dx}{x^2(x+1)^3}$ 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.
    $\displaystyle f(x)$ is quadratic such that $\displaystyle f(0) = 1$ ...

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

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

    $\displaystyle f'(0) = b$


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

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

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

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

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

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

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

    equating coefficients ...

    $\displaystyle B+D = 0$

    $\displaystyle 3B+D+E = a$

    $\displaystyle 3B = b$

    $\displaystyle B = 1$

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