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Math Help - 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 \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 \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<br />

    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<br />

    equating coefficients ...

    B+D = 0

    3B+D+E = a

    3B = b

    B = 1

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