Answer
$$\displaystyle{\int_{\frac{\sqrt2}{3}}^{\frac{2}{3}}\frac{1}{x^5\sqrt{9x^2-1}}dx=\frac{567\sqrt3}{64}-\frac{81}{4}+\frac{81}{32}\pi}\\$$
Work Step by Step
$\displaystyle{I=\int_{\frac{\sqrt2}{3}}^{\frac{2}{3}}\frac{1}{x^5\sqrt{9x^2-1}}dx}\\$
$\displaystyle \left[\begin{array}{ll} 3x=\sec\theta & 9x^2=\sec^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{1}{3}\sec\theta\tan\theta & dx=\frac{1}{3}\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{1}{\frac{1}{243}\sec^5\theta\sqrt{\sec^2\theta-1}}\times\frac{1}{3}\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{81}{\sec^4\theta\sqrt{\tan^2\theta}}\tan\theta\ d\theta}\\ \displaystyle{I=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{81}{\sec^4\theta}\ d\theta}\\ \displaystyle{I=81\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\cos^4\theta\ d\theta}\\ \displaystyle{I=81\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left(\cos^2\theta\right)^2\ d\theta}\\ \displaystyle{I=81\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left(\frac{1}{2}\left(\cos2\theta+1\right)\right)^2\ d\theta}\\ \displaystyle{I=81\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{1}{4}\left({\cos^22\theta}+2\cos2\theta+1\right)\ d\theta}\\ \displaystyle{I=81\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{1}{4}\left(\frac{\cos4\theta+1}{2}+2\cos2\theta+1\right)\ d\theta}\\ \displaystyle{I=\frac{81}{8}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left({\cos4\theta}+4\cos2\theta+3\right)\ d\theta}\\ \displaystyle{I=\frac{81}{8}\left[{\frac{1}{4}\sin4\theta}+2\sin2\theta+3\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}}\\ \displaystyle{I=\frac{81}{32}\left[{\sin4\theta}+8\sin2\theta+12\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}}\\ \displaystyle{I=\frac{81}{32}\left(-\frac{\sqrt3}{2}+4\sqrt3+4\pi\right)-\frac{81}{32}\left(0+8+3\pi\right)}\\ \displaystyle{I=\frac{567\sqrt3}{64}+\frac{81}{8}\pi-\frac{81}{4}-\frac{243}{32}\pi}\\ \displaystyle{I=\frac{567\sqrt3}{64}-\frac{81}{4}+\frac{81}{32}\pi}\\$