1. **State the problem:** Evaluate the integral $$\int \frac{x}{\sqrt{81x^2 - 1}} \, dx$$. 2. **Identify the integral type:** This is an integral involving a rational function with a square root of a quadratic expression in the denominator. 3. **Use substitution:** Let $$u = 81x^2 - 1$$. Then, $$\frac{du}{dx} = 162x$$, so $$du = 162x \, dx$$. 4. **Rewrite the integral:** We have $$x \, dx = \frac{du}{162}$$, so the integral becomes $$\int \frac{x}{\sqrt{81x^2 - 1}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{162} = \frac{1}{162} \int u^{-\frac{1}{2}} \, du$$. 5. **Integrate:** Using the power rule for integrals, $$\int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} + C$$. 6. **Substitute back:** So, $$\frac{1}{162} \int u^{-\frac{1}{2}} \, du = \frac{1}{162} \cdot 2u^{\frac{1}{2}} + C = \frac{2}{162} \sqrt{81x^2 - 1} + C = \frac{1}{81} \sqrt{81x^2 - 1} + C$$. **Final answer:** $$\int \frac{x}{\sqrt{81x^2 - 1}} \, dx = \frac{1}{81} \sqrt{81x^2 - 1} + C$$.