## Calculus (3rd Edition)

The integral converges and its value is $\frac{5}{7}\frac{1}{9^{7/5}}$.
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 $\frac{5}{7}\frac{1}{9^{7/5}}$.