Integral inequality 5 How can I prove that:
$$8\le \int _3^4\frac{x^2}{x-2}dx\le 9$$
My teacher advised me to find the asymptotes, why? what helps me if I find the asymptotes?

Direct approach is not so painful:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx = \int_{1}^{2}\frac{(x+2)^2}{x}\,dx = \int_{1}^{2}\left(x+4+\frac{4}{x}\right)\,dx =\frac{11}{2}+4\log 2 $$
and by the Hermite-Hadamard inequality we have:
$$ \log 2 =\int_{1}^{2}\frac{dx}{x} \in \left(\frac{2}{3},\frac{3}{4}\right) $$
so:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx \in \left(8+\frac{1}{6},8+\frac{1}{2}\right).$$
