If $f$ is quadratic function such that $f(0)=1$ and $\int\frac{f(x)}{x^2(x+1)^3}dx $ is a rational function, find the value of $f'(0)$.
I already tried solving this question by using general quadratic equation $ax^2+bx+c$ and then using partial fraction method but it became very complicated.
We have that $b=f'(0)$ and $c=f(0)=1$. Moreover, by using partial fraction method we get $$\frac{ax^2+bx+1}{x^2(x+1)^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{(x+1)^3}.$$ If the integral is a rational function, i.e. a ratio of polynomials, then $A=0$ and $C=0$ because the integrals of those terms yield logarithms. Hence $$ax^2+bx+1=B(x+1)^3+Dx^2(x+1)+Ex^2,$$ that is $$(B+D)x^3+(3B+D+E-a)x^2+(3B-b)x+ (B-1)=0.$$ Are you able to find $b$?