Your attachment doesn't have anything to do with this question! Yes, a quadratic polynomial, P(x), such that P(0)= 1 and P'(0)= 0, is of the form P(x)= ax^2+ 1. The third condition is that "the integral of P(x) over x^3(x-1)^2 is a rational function".
That is $\displaystyle \int \frac{ax^2+ 1}{x^3(x- 1)^2}dx$. Do that integration by "partial fractions". We can write $\displaystyle \frac{ax^2+ 1}{x^3(x- 1)^2}$ as $\displaystyle \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x^3}+ \frac{D}{x- 1}+ \frac{E}{(x- 1)^2}$.
The crucial point is that the integrals of $\displaystyle \frac{A}{x}$ and $\displaystyle \frac{D}{x- 1}$ are logarithms, not rational functions. Can you find "a" such that A and D are 0?