Answer
The integral converges and its value is $\frac{5}{7}\frac{1}{9^{7/5}}$.
Work Step by Step
We have
$$
\int_{9}^{\infty} \frac{d x}{x^{12/5}}= \int_{9}^{\infty}x^{-12/5}dx\\
=-\frac{1}{7/5}x^{-7/5}|_9^{\infty}=-\frac{5}{7}(\frac{1}{\infty}-\frac{1}{9^{7/5}})=\frac{5}{7}\frac{1}{9^{7/5}}
$$
Hence, the integral converges and its value is $\frac{5}{7}\frac{1}{9^{7/5}}$.
