1. **State the problem:** Evaluate the integral $$\int \frac{1}{x^2 \sqrt{4x^2 - 9}} \, dx$$. 2. **Identify the form and substitution:** The integrand contains a square root of the form $$\sqrt{a^2 x^2 - b^2}$$ with $$a=2$$ and $$b=3/2$$. A common substitution for integrals involving $$\sqrt{a^2 x^2 - b^2}$$ is $$x = \frac{b}{a} \sec \theta$$. 3. **Apply substitution:** Let $$x = \frac{3}{2} \sec \theta$$, then $$dx = \frac{3}{2} \sec \theta \tan \theta \, d\theta$$. 4. **Rewrite the integral:** Substitute into the integral: $$\sqrt{4x^2 - 9} = \sqrt{4 \left(\frac{3}{2} \sec \theta\right)^2 - 9} = \sqrt{9 \sec^2 \theta - 9} = 3 \sqrt{\sec^2 \theta - 1} = 3 \tan \theta$$. Also, $$x^2 = \left(\frac{3}{2} \sec \theta\right)^2 = \frac{9}{4} \sec^2 \theta$$. So the integrand becomes: $$\frac{1}{x^2 \sqrt{4x^2 - 9}} = \frac{1}{\frac{9}{4} \sec^2 \theta \cdot 3 \tan \theta} = \frac{1}{\frac{27}{4} \sec^2 \theta \tan \theta} = \frac{4}{27} \cdot \frac{1}{\sec^2 \theta \tan \theta}$$. 5. **Substitute dx:** $$dx = \frac{3}{2} \sec \theta \tan \theta \, d\theta$$. 6. **Integral in terms of $$\theta$$:** $$\int \frac{1}{x^2 \sqrt{4x^2 - 9}} \, dx = \int \frac{4}{27} \cdot \frac{1}{\sec^2 \theta \tan \theta} \cdot \frac{3}{2} \sec \theta \tan \theta \, d\theta = \int \frac{4}{27} \cdot \frac{3}{2} \cdot \frac{\sec \theta \tan \theta}{\sec^2 \theta \tan \theta} \, d\theta$$ Simplify the fraction inside the integral: $$\frac{\sec \theta \tan \theta}{\sec^2 \theta \tan \theta} = \frac{1}{\sec \theta} = \cos \theta$$. So the integral becomes: $$\int \frac{4}{27} \cdot \frac{3}{2} \cos \theta \, d\theta = \int \frac{2}{9} \cos \theta \, d\theta$$. 7. **Integrate:** $$\int \frac{2}{9} \cos \theta \, d\theta = \frac{2}{9} \sin \theta + C$$. 8. **Back-substitute $$\theta$$:** Recall $$x = \frac{3}{2} \sec \theta$$, so $$\sec \theta = \frac{2x}{3} \implies \cos \theta = \frac{3}{2x}$$. Since $$\sin^2 \theta + \cos^2 \theta = 1$$, $$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{3}{2x}\right)^2} = \sqrt{1 - \frac{9}{4x^2}} = \sqrt{\frac{4x^2 - 9}{4x^2}} = \frac{\sqrt{4x^2 - 9}}{2x}$$. 9. **Final answer:** $$\int \frac{1}{x^2 \sqrt{4x^2 - 9}} \, dx = \frac{2}{9} \cdot \frac{\sqrt{4x^2 - 9}}{2x} + C = \frac{\sqrt{4x^2 - 9}}{9x} + C$$. **Answer:** $$\boxed{\frac{\sqrt{4x^2 - 9}}{9x} + C}$$