<s>[INST] If $\int \frac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is rational and $f$ is quadratic with $f(0)=1$, then find $f'(0)$ 
If $\int \dfrac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is a rational function, and $f$ is quadratic function, such that $f(0)=1$. Then Find $f'(0)$

This looks like an interesting problem with an elegant solution.
But I cannot figure out how to approach this.
Can anyone help
 [/INST] Let $f(x):=ax^2+bx+1$
$$\int \frac{ax^2+bx+1}{x^2(x+1)^3}dx=\int\left(\frac{a-b+1}{(x+1)^3}+\frac{2-b}{(x+1)^2}+\frac{b-3}{x}+\frac{3-b}{x+1}+\frac1{x^2}\right)dx$$
So $b=3$
Now $f'(x)=2ax+b,f'(0)=b=3$
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