Finding the value of $f'(0)$ 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, how can we find the value of $f'(0)$? I am totally clueless to this. Any tip on how to start? If you wish to give details, then many thanks to you.

Note that
$${f(x)\over x^2(x+1)^3}={xf(x)\over(x^2+x)^3}$$
Let $u=x^2+x$.  Suppose 
$$xf(x)=u{du\over dx}=u(2x+1)=(x^2+x)(2x+1)=2x^3+3x^2+x=x(2x^2+3x+1)$$
Then
$$\int{f(x)\over x^2(x+1)^3}dx=\int{xf(x)\over(x^2+x)^3}dx=\int{udu\over u^3}=\int{du\over u^2}=-{1\over u}+C=-{1\over(x^2+x)}+C$$
is a rational function. So $f(x)=2x^2+3x+1$ does the trick, and $f'(0)=3$.
