$$\int_9^{\infty} \frac{1}{\sqrt{x^5}} dx$$
$$u=\sqrt{x^5}\implies u=x^{5/2} \implies x=u^{2/5}\implies dx=\frac{2}{5u^{3/5}}$$
$$\int_{9^{5/2}}^{\infty} \frac{2}{5u^{3/5}u} du$$
$$\int_{9^{5/2}}^{\infty} \frac{2}{5u^{8/5}} du$$
$$\int_{9^{5/2}}^{\infty} \frac{2u^{-8/5}}{5} du$$
$$\left[ \frac{2u^{-3/5}}{-3} \right]_{9^{5/2}}^{\infty}$$
$$=0-\left(-\frac{2\cdot 9^{-3/2}}{-3}\right)$$
Is this correct?
Any faster method suggestions?
What you did works, though you threw in an extra minus sign at the end so your answer came out negative when it should be positive.
But you don't need the substitution, it's way overkill. $$\frac{1}{\sqrt{x^5}} = x^{-5/2}$$ $$\int_9^\infty x^{-5/2} = -\frac{2}{3}x^{-3/2}\big|_9^\infty = 0 - \left(-\frac{2}{3}9^{-3/2}\right) = \frac{2}{81}$$