Step-by-step Solution:

To find the new quadratic equation, we first need to determine the sum and product of its roots. A new quadratic equation with roots \(R_1\) and \(R_2\) can be written as \(x^2 - (R_1+R_2)x + (R_1R_2) = 0\).

1. Use Vieta's formulas for the original equation:
For the equation \(x^2 + ax + b = 0\) with roots \(\alpha\) and \(\beta\):
- Sum of roots: \(\alpha + \beta = -a\)
- Product of roots: \(\alpha \beta = b\)

2. Define the new roots and find their sum (S') and product (P'):
The new roots are \(R_1 = \frac{1}{\alpha^2+\alpha}\) and \(R_2 = \frac{1}{\beta^2+\beta}\).

Product of new roots (P'):
\[P' = R_1 R_2 = \frac{1}{(\alpha^2+\alpha)(\beta^2+\beta)} = \frac{1}{\alpha^2\beta^2 + \alpha^2\beta + \alpha\beta^2 + \alpha\beta}\]\[P' = \frac{1}{(\alpha\beta)^2 + \alpha\beta(\alpha+\beta) + \alpha\beta}\]Substituting \(\alpha+\beta = -a\) and \(\alpha\beta = b\):
\[P' = \frac{1}{b^2 + b(-a) + b} = \frac{1}{b^2 - ab + b}\]
Sum of new roots (S'):
\[S' = R_1 + R_2 = \frac{1}{\alpha^2+\alpha} + \frac{1}{\beta^2+\beta} = \frac{(\beta^2+\beta) + (\alpha^2+\alpha)}{(\alpha^2+\alpha)(\beta^2+\beta)}\]The denominator is the inverse of \(P'\). The numerator is:
\[(\alpha^2+\beta^2) + (\alpha+\beta)\]We know \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-a)^2 - 2b = a^2-2b\).
So, the numerator is \((a^2-2b) + (-a) = a^2-a-2b\).
Therefore, the sum of the new roots is:
\[S' = \frac{a^2-a-2b}{b^2-ab+b}\]
3. Form the new quadratic equation:
The new equation is \(x^2 - S'x + P' = 0\).
\[x^2 - \left( \frac{a^2-a-2b}{b^2-ab+b} \right)x + \frac{1}{b^2-ab+b} = 0\]To clear the fractions, we multiply the entire equation by the common denominator \((b^2-ab+b)\):
\[(b^2-ab+b)x^2 - (a^2-a-2b)x + 1 = 0\]
4. Analysis of Options:
The correctly derived equation is \((b^2-ab+b)x^2 - (a^2-a-2b)x + 1 = 0\). This result does not match any of the provided multiple-choice options. The complexity and structure of the options (e.g., terms like \(a^3\) and \(b^3\)) suggest that there is a significant error in the problem statement or the options as provided.