Question:

$\because 0 < x^3 < x^2$, for all $x \in (0,1) \Rightarrow x^2 < x^2 + x^3 < 2x^2 \Rightarrow -2x^2 < -x^2 - x^3 < -x^2 \Rightarrow 9 - 2x^2 < 9 - x^2 - x^3 < 9 - x^2 \Rightarrow \frac{1}{9-x^2} < \frac{1}{9-x^2-x^3} < \frac{1}{9-2x^2} \Rightarrow f_1(x) < f_3(x) < f_2(x)$ for $x \in (0, 1) \Rightarrow \int_{0}^{1} f_1(x) dx < \int_{0}^{1} f_3(x) dx < \int_{0}^{1} f_2(x) dx \Rightarrow \int_{0}^{1} \frac{dx}{9-x^2} < \int_{0}^{1} \frac{dx}{9-x^2-x^3} < \int_{0}^{1} \frac{dx}{9-2x^2} \Rightarrow \frac{1}{6} \left( \ln \left| \frac{3+x}{3-x} \right| \right)_0^1 < \int_{0}^{1} \frac{dx}{9-x^2-x^3} < \frac{1}{6\sqrt{2}} \left( \ln \left| \frac{3+2x}{3-2x} \right| \right)_0^1 \Rightarrow \frac{1}{6} \ln 2 < \int_{0}^{1} \frac{1}{9-x^2-x^3} dx < \frac{1}{6\sqrt{2}} \ln 5$