Finding the value of $f'(0)$

Let $f(x) = ax^2 + bx + c.$ As $f(0) = 1,$ we have that $f(x) = ax^2 + bx + 1.$ Then $$I=\int \frac{f(x) dx}{x^2(x+1)^3} = \int \frac{a}{(x+1)^3}+\frac{b}{x(x+1)^3}+\frac{1}{x^2(x+1)^3}\,dx \\\stackrel{Partial Fractions}{=} -\frac 1x + \frac{b-a-1}{2 (1 + x)^2} + \frac{-2 + b}{1 + x} + (b-3)\log(x) - (b-3)\log(1 + x) + C.$$ As the function must be rational, we must have $b=3,$ which leads $f'(0) = \left.2ax+b\right|_{x=0} = \boxed{3}$

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 $f'(0)=3$.